7:20 at this point, rather than work out the arithmetic, I would much prefer just saying: these correspond to angles of +/- 60 and 30 degrees respectively, but dont forget +/- 120 and 150 degrees (draw 1 diagram and remind where the sin is on the unit circle). The 2 solutions make intuitive sense because of the sin/cos symmetry in the equation. That would be much more satisfying than the slow arithmetic imo.
Misalkan a:=81^sin^2x & b:=81^cos^2x, maka .loga:=sin^2x log81; sin^2x= loga/log81 & cos^2x=logb/log81. Dan sin^2x+cos^x=1 yg berarti ( loga)+(logb)=log81 .ab=81 Serta a+b=30 sehingga a=27&b=3
Nice combinaison if two solving ideas. First, usage of the trigonometric formula then transformation into a quadratic equation. Nothing complex at the end while the initial equation looks scary. 😅 Maths is also about have the right idea at the right time, keep your nerve and continue trying ! Thanks for the video !
I would like to see more Math Olympiad problems using trigonometry as in my youth I participated in my country math olympiad where I was asked to resolve trigonometry equations and complex number problems. For instance, prove that: (cos α + i sin α)(cos β + i sin β) = cos (α + β) + i sin (α + β) o ( cos α + i sin α )ⁿ = cos nα + i sin nα Thank you for your videos.
To prove the first given equation: (cos α + i sin α)(cos β + i sin β) = cos (α + β) + i sin (α + β), we can expand the left-hand side (LHS) using the properties of complex numbers: LHS = cos α * cos β + i sin α * cos β + i cos α * sin β + i * i * sin α * sin β. Remembering that i * i is equal to -1, we can simplify the expression: LHS = cos α * cos β + i sin α * cos β + i cos α * sin β - sin α * sin β. Now, let's use the trigonometric identity: cos (α + β) = cos α * cos β - sin α * sin β, sin (α + β) = sin α * cos β + cos α * sin β. Comparing these identities with the simplified LHS, we can see that: cos (α + β) + i sin (α + β) = cos α * cos β + i sin α * cos β + i cos α * sin β - sin α * sin β, which is equal to the simplified LHS. Therefore, we have proven that: (cos α + i sin α)(cos β + i sin β) = cos (α + β) + i sin (α + β). To prove the second one: (cos α + i sin α)ⁿ = cos nα + i sin nα, we can use Euler's formula, which states: e^(iθ) = cos θ + i sin θ. By substituting θ = α into Euler's formula, we get: e^(iα) = cos α + i sin α. Now, let's raise both sides of the equation to the power of n: [e^(iα)]ⁿ = (cos α + i sin α)ⁿ. Using the properties of exponentiation, we can expand the left-hand side (LHS): [e^(iα)]ⁿ = e^(iα * n) = e^(i(nα)). By Euler's formula, e^(i(nα)) can be expressed as: e^(i(nα)) = cos(nα) + i sin(nα).
yes, but no 1/3Pi+2kPi and 4/3Pi+2kPi can be merged to 1/3Pi+kPi because 4/3=1+1/3 other pairs are 2/3 and 5/3; 1/6 and 7/6; 5/6 and 11/6 so it is 1/6Pi; 1/3Pi; 2/3Pi; 5/6Pi whit kPi periodicity
@@matteopriotto5131 yes, but no. 2pi periodicity comes because the original equation uses sinus and cosinus, functions whit 2pi perioditicty. my solution uses pi periodicity because sine squared (and cosine squared) has pi periodicity. so pi and 2pi are general solotions for this type of equations. 1/2pi solution comes because of the specific numbers used in the equation to be solved without a calculator. not a general solution. so, i would not recommend using it. but, you have a really good eye, i didn't notice it. :)
@@nagydee yeah but the function y=81^(sin²x)+81^(cos²x) does have pi/2 periodicity, just replace x with x+pi/2 and you get the same thing. I actually noticed this only now that you made me think about it, when I solved the problem I just thought "wow the solutions repeat every pi/2, cool" and that was it. I know everything you said about periodicity of goniometric functions, but I don't see the problem with changing the periodicity in the solution accordingly if you notice that there's some kind of pattern. It's not wrong if you don't of course, but neither if you do.
All the solutions can be written compactly as (5pi)/6 + (pi)k. When k is even you get all of your first solutions, when k is odd you get all of your second solutions.
It’s right, but the value root3/2 which is noticed along the dotted line can be confused with a ccoordinate on the cosinus axis, althought it’s really a coordinate on sinus axis like the author think it well
Just write 81^cos^2x= 81^1-sin^2x.then let 81^sin^2x=P.then the whole equation will become a quadratic in p.Solve and get p and then simultaneously x....
Noting that the set of solutions is all those angles which are multiples of pi/6 but are not zero or multiples of pi/2, this means the solutions can be written as pi/6 + k pi/2 and pi/3 + k pi/2, for k an integer. Interestingly, advancing around the circle in pi/2 steps alternates between the cases sin(x)^2 =1/4 and = 3/4.
La façon suivie pour résoudre ce problème est correcte.mais la dernière étape est fausse.la maîtresse a calculé cos x=1/2. Le sin x se fait sur l'autre axe😅😂😊
Thnx for the vid. Interesting solution. However IMO it's a bit long-winded; the 4 possible solutions could have been derived much faster with no need of drawing any unit circles etc.: Refactor the original equation as a quadratic (where your unknown variable has either the sin squared or cos squared in the exponent - doesn't matter which, just choose one). Solve the quadratic then take logs of both sides of the resulting equations and finally solve for the angles "x". 30° , 60°, 120°, 150°
Well explained, but IMO, when you are solving equations with trig functions as exponents, you may assume some prior knowledge in your viewers and take somewhat bigger steps.
What is K? (K는 자연수)안 써도 돼는 어떤 약속된 k인가? K가 무슨 용어야? 뭐 요즘 뭐 바꼈냐? 새로 생긴거임? 요즘 학교서 이렇게 배우는데 내가 나이먹어서 모르는 거임? 저 나라에서만 통용되는 k인가? K가 자연수인게 필요 없나요? Constance 인지 constant 인지 라고 말했으니까 K에대한 어떤 범위,조건이 있지 않아야 할까? K는 뭘까? 쌤 ! K에 아무거나 막 넣어도 돼요? 1/2 넣어 볼까요? 어때요?
I agree, it seems to be a mixture of 1st year (year 7), 3rd/4th year (year 9/10) and 6th form (year 12) techniques. (It's hard to say exactly what years as when things are taught have changed since my school days.)
These are nothing 😂😂....i had tackled many of these types of problems in my 12th class and the most funny thing was that i am not a topper type of guy ,am from that of the avg one .In India these questions came to be knows as the simplest one 😂😂
Not sure what you mean. She's using a unit circle and the dotted lines are of length of the sines - the y component of the point on the circle - which she uses. Going back to trig ratios and degrees is what confused me. This type of question is going to be used when you've met: sin x = x - ×^3/3! + ×^5/5! - x^7/7! + ... cos x = 1 - ×^2/2! + x^4/4! - x^6/6! + ... where x is (has to be) measured in radians.
Олимпиадная зажача для класса 8-9? Сдишком уж лчевидное решение, можно было бы в лоб решать без замены, так как корни выдны сразу же. Обидно, что наше общество так сильно загнивает при капитализме...
@@irvingrabin I think he means taking log to the base 81 of each side, giving sin^2 (x) + cos^2 (x) = log,81(30), use the double angle formula of cos to create an expression in terms of sin, let u = sin(x), solve for u, substitute and find your answer that way
@@ow3nthehorror Incorrect usage of logarithmic rules aside, that method of finding x wouldn't even work considering you're essentially writing the equation as 1 = log81(30), meaning there's no x that'd satisfy it no matter what.
هده هي المعادلة تربط الاداتين المعياري الدي عير بمعيار لا علاقة له بزمان او مكان .، يرتبطان المعايرين بذات واهية لا وجود لها اصلا .، نحدد الاكس بدات المعيار من كوسينهات و السينيسهات فقط .،
You overcomplicated this poor little thing. Here is how to destroy it. First of all I'm introducing y=sin²x and noticing that cos²x=1-sin²x our equation becomes: 81^y+81^(1-y)=30 Not very beautiful. Let's divide by 9... 81^[y-1/2]+81^[1/2-y]=10/3. Much better! Let's introduce z=81^[y-1/2]. Our equation becomes: z+1/z=10/3. We can see an obviouss solution, which is 3: 10/3=3+1/3. And since z and 1/z are symetrical that means that 1/3 is also a solution. But if we multiply our equation by z, we obtain a quadratic equation and we know that we have at most two solutions. We found two solutions, we're done. So now we have to solve 81^[y-1/2]=3 or 81^[y-1/2]=1/3 Let's rewrite this as powers of 3: 3^[4(y-1/2)]=3 or 3^[4(y-1/2)]=3^(-1) Then 4y-2=1 or 4y-2=-1 Then y=3/4 or y=-1/4. But y=sin²x which means that the possible values of sin(x) are sqrt(3)/2, -sqrt(3)/2, 1/2 or -1/2 Then the possible values of x are pi/6, pi/3, 2.pi/3, 5.pi/6 and their opposites.
hmm I don't know what 'quick' method everyone else is saying. but I did have my own method which was quicker than this. you begin by re-writing 81 as 3^4 and then multiplying the exponentials of both terms. Next, I realised that I could rewrite 30 as 3^3 + 3^1. This meant that I could equate the exponential so that when 4Sin^2x = 3, 4Cos^2x = 1. Then I multiplied the cosine function by 3 to get 12Cos^2x = 3 and then equated the sin and cosine function. Rearranging you would get that tan^2x = 3 meaning tanx = root3. Knowing your exact values, you should be able to tell that x = + or - 60 degrees otherwise +- pi/3 radians. Now i look back at the original equation and notice that sin(pi - x) = sin(x) and so 2pi/3 is also an answer. Next to find other solutions is when 4Sin^2x = 1 then 4cos^2x = 3. using the same method, you get tanx = 1/root3 which means x = 30 degrees or +-pi/6. from the solutions of +-pi/3 and +-pi/6 and 2pi/3 is also an answer you find that for each base angle of pi/3 or pi/6, there are solutions which repeat every pi/2.
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write 81^sin^2 (x) as y and because cos^2 (x) = 1 - sin^2 (x), then write 81^cos^2 (x) as 81/y. this becomes a quadratic and continue from there.
7:20 at this point, rather than work out the arithmetic, I would much prefer just saying: these correspond to angles of +/- 60 and 30 degrees respectively, but dont forget +/- 120 and 150 degrees (draw 1 diagram and remind where the sin is on the unit circle). The 2 solutions make intuitive sense because of the sin/cos symmetry in the equation. That would be much more satisfying than the slow arithmetic imo.
Misalkan a:=81^sin^2x & b:=81^cos^2x, maka
.loga:=sin^2x log81; sin^2x= loga/log81 & cos^2x=logb/log81.
Dan sin^2x+cos^x=1 yg berarti
( loga)+(logb)=log81
.ab=81
Serta a+b=30 sehingga a=27&b=3
Nice combinaison if two solving ideas. First, usage of the trigonometric formula then transformation into a quadratic equation. Nothing complex at the end while the initial equation looks scary. 😅 Maths is also about have the right idea at the right time, keep your nerve and continue trying ! Thanks for the video !
LK, you are clever and a good presenter, and totally adorable ❤❤❤
I liked this exercise so much. Thanks for sharing.
Nicely done. I wouldn't have bothered to write 81 as 3^4
I would like to see more Math Olympiad problems using trigonometry as in my youth I participated in my country math olympiad where I was asked to resolve trigonometry equations and complex number problems. For instance, prove that:
(cos α + i sin α)(cos β + i sin β) = cos (α + β) + i sin (α + β)
o
( cos α + i sin α )ⁿ = cos nα + i sin nα
Thank you for your videos.
Complex numbers and exponential
To prove the first given equation:
(cos α + i sin α)(cos β + i sin β) = cos (α + β) + i sin (α + β),
we can expand the left-hand side (LHS) using the properties of complex numbers:
LHS = cos α * cos β + i sin α * cos β + i cos α * sin β + i * i * sin α * sin β.
Remembering that i * i is equal to -1, we can simplify the expression:
LHS = cos α * cos β + i sin α * cos β + i cos α * sin β - sin α * sin β.
Now, let's use the trigonometric identity:
cos (α + β) = cos α * cos β - sin α * sin β,
sin (α + β) = sin α * cos β + cos α * sin β.
Comparing these identities with the simplified LHS, we can see that:
cos (α + β) + i sin (α + β) = cos α * cos β + i sin α * cos β + i cos α * sin β - sin α * sin β,
which is equal to the simplified LHS. Therefore, we have proven that:
(cos α + i sin α)(cos β + i sin β) = cos (α + β) + i sin (α + β).
To prove the second one:
(cos α + i sin α)ⁿ = cos nα + i sin nα,
we can use Euler's formula, which states:
e^(iθ) = cos θ + i sin θ.
By substituting θ = α into Euler's formula, we get:
e^(iα) = cos α + i sin α.
Now, let's raise both sides of the equation to the power of n:
[e^(iα)]ⁿ = (cos α + i sin α)ⁿ.
Using the properties of exponentiation, we can expand the left-hand side (LHS):
[e^(iα)]ⁿ = e^(iα * n) = e^(i(nα)).
By Euler's formula, e^(i(nα)) can be expressed as:
e^(i(nα)) = cos(nα) + i sin(nα).
By inducyion method easy
Why not substitute u=... much before?? For instance, I started doing so and it made things much easier from the start, reducing the risk of mistakes.
Nice piece of work. Thanks a lot for sharing nice stuff ❤❤❤
yes, but no
1/3Pi+2kPi and 4/3Pi+2kPi can be merged to 1/3Pi+kPi because 4/3=1+1/3
other pairs are 2/3 and 5/3; 1/6 and 7/6; 5/6 and 11/6
so it is
1/6Pi; 1/3Pi; 2/3Pi; 5/6Pi whit kPi periodicity
aka ⅙pi; ⅓pi with pi/2 periodicity
@@matteopriotto5131 yes, but no.
2pi periodicity comes because the original equation uses sinus and cosinus, functions whit 2pi perioditicty. my solution uses pi periodicity because sine squared (and cosine squared) has pi periodicity.
so pi and 2pi are general solotions for this type of equations.
1/2pi solution comes because of the specific numbers used in the equation to be solved without a calculator. not a general solution.
so, i would not recommend using it.
but, you have a really good eye, i didn't notice it. :)
@@nagydee yeah but the function y=81^(sin²x)+81^(cos²x) does have pi/2 periodicity, just replace x with x+pi/2 and you get the same thing. I actually noticed this only now that you made me think about it, when I solved the problem I just thought "wow the solutions repeat every pi/2, cool" and that was it. I know everything you said about periodicity of goniometric functions, but I don't see the problem with changing the periodicity in the solution accordingly if you notice that there's some kind of pattern. It's not wrong if you don't of course, but neither if you do.
All the solutions can be written compactly as (5pi)/6 + (pi)k. When k is even you get all of your first solutions, when k is odd you get all of your second solutions.
It’s right, but the value root3/2 which is noticed along the dotted line can be confused with a ccoordinate on the cosinus axis, althought it’s really a coordinate on sinus axis like the author think it well
Just write 81^cos^2x= 81^1-sin^2x.then let 81^sin^2x=P.then the whole equation will become a quadratic in p.Solve and get p and then simultaneously x....
Selanjutnya tentu
sin^2x=log27/log81=3/4; sinx=¥(3/4)
.x=60°(-1)^n+n.180°
Juga cos^2x=log3/log81=1/4
.cosx÷1/2; x=+_(60°)+n.360°
Ingat pers trigonometri melibatkan sifat periodik sinusoia
Noting that the set of solutions is all those angles which are multiples of pi/6 but are not zero or multiples of pi/2, this means the solutions can be written as pi/6 + k pi/2 and pi/3 + k pi/2, for k an integer. Interestingly, advancing around the circle in pi/2 steps alternates between the cases sin(x)^2 =1/4 and = 3/4.
Very nicely explained simple solution. very good.
According to Hannah Arendt self-reflection needs to start based upon own experience in order to deliver positive results.
Very well explained
Use the satndard solution of trig eq
Cos²theta= cos ²a
=> Theta =. nπ +or - a
تمرين جميل جيد. اللهم شرح واضح مرتب. شكرا جزيلا لكم والله يحفظكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .
ما ينفع تدعي لكافرة بأن الله يحفظها(الا اذا كانت مسلمة)
The pace is quite good; keeping others from skipping steps and getting errors.
La façon suivie pour résoudre ce problème est correcte.mais la dernière étape est fausse.la maîtresse a calculé cos x=1/2. Le sin x se fait sur l'autre axe😅😂😊
I think that's a bit slow
Thnx for the vid. Interesting solution. However IMO it's a bit long-winded; the 4 possible solutions could have been derived much faster with no need of drawing any unit circles etc.:
Refactor the original equation as a quadratic (where your unknown variable has either the sin squared or cos squared in the exponent - doesn't matter which, just choose one). Solve the quadratic then take logs of both sides of the resulting equations and finally solve for the angles "x". 30° , 60°, 120°, 150°
Shouldn't the answer be x = nπ + (-1)^(n).α or, nπ + (-1)^(n).β where n∈ℤ; α= +π/3 or -π/3; β = +π/6 or -π/6 ?????
Oh can u upload some hard problems from IMO please edit:these are very easy sums
Aapka maths samjhne ke liye maths expert hone chahiye......
As for me, it was a really easy task for Olympiad.
Well that was fun!😮
Amazing- thanks
Excelente.
nicely done
Thanks, miss.
Well explained, but IMO, when you are solving equations with trig functions as exponents, you may assume some prior knowledge in your viewers and take somewhat bigger steps.
I fully agree with your bigger steps point. She is also a good presenter, as you say.
Интересная задача получится, если слева будет не три корня последовательно, а бесконечное количество корней..
What is K?
(K는 자연수)안 써도 돼는 어떤 약속된 k인가?
K가 무슨 용어야?
뭐 요즘 뭐 바꼈냐?
새로 생긴거임?
요즘 학교서 이렇게 배우는데 내가 나이먹어서 모르는 거임?
저 나라에서만 통용되는 k인가?
K가 자연수인게 필요 없나요?
Constance 인지 constant 인지 라고 말했으니까
K에대한 어떤 범위,조건이 있지 않아야 할까?
K는 뭘까?
쌤 ! K에 아무거나 막 넣어도 돼요?
1/2 넣어 볼까요?
어때요?
Excellent
Why doesn’t the 120 degrees and 60 degrees cover a straight line
Nice, but y didn't u finish your work like expressing the both values of x magnitudinally like so the value of x here is x = xxx ans like that.
We can do it more easily and faster by substitution
Mental arithmetic, i got two possible: 30° and 60°， both are fit
Es bellísimo ❤❤❤❤
Thanks for this solution , but for some basic algebra always draw an assumption that at this level it's expected to be equiped with those basics
I agree, it seems to be a mixture of 1st year (year 7), 3rd/4th year (year 9/10) and 6th form (year 12) techniques.
(It's hard to say exactly what years as when things are taught have changed since my school days.)
Not relevant to your comment, but solving this by taking log on both sides would have been much faster
@@civilbro5800 how do you evaluate:
log (b^x *+* b^y)
Useful mathematics
이걸보면 내가 옛날에 수학시간에 앞에서 설명을 하는게 두려웠다. 나는 아직도 수학이 두렵다. 이걸 보면서 나에 뜨라우마가 생겼다.
These are nothing 😂😂....i had tackled many of these types of problems in my 12th class and the most funny thing was that i am not a topper type of guy ,am from that of the avg one .In India these questions came to be knows as the simplest one 😂😂
I was thinking the samw
I did it even without knowing about trigonometric equations but only basics of class 10 knowledge
I often take log in such situations though
Can't we use AM GM in equality
Nice problem ❤
What a sweet voice
Bruh you lost me at the coordinate plane
Great
Excelente
Super 🎉
Don't you need a statement on "K" being integer?
나더러 틀렸는데
왜 틀렸다 안 하냐고
그건 제가 아니죠
당신들 짓이지
쌤, 재밌어요
계속 영상 업로드 해주세요
지금부터 영상이 틀렸는데 왜 틀렸다 안 하냐 나한테 시비걸면 무시
Its not maths, its like puzzles.
Le sinus x se trouve sur l'axe des ordonnés pas sur l'axe des abscisses
Corriger votre vidéo
I like it.
acrsin(-sqrt(3)/2) = -60 degrees, not 120. Just picture the graph of sin x in your head - or use a calculator.
ola boa noite! parabens pelo video
muito legal gostei muito shwou de bola valeu
А почему синус проекция на ось Х а не на ось У?
x is 30 degree or 60 degree. 4 minute job 😊😊
Math Olympiad Question 😅 I think this problem is far from Olympiad question. 😀
بما سينسكس من الاعشار و الكوسنكس من الآلاف كحد أقصى المعطيات و المصطلح المعياري .،فإن =
Sinx= root 3 / 2😊
81 sin 2x+81 cos2x=30
Simple Pure 2 question
Degrees are correct but, but sinus on the y axis
Not sure what you mean. She's using a unit circle and the dotted lines are of length of the sines - the y component of the point on the circle - which she uses.
Going back to trig ratios and degrees is what confused me. This type of question is going to be used when you've met:
sin x = x - ×^3/3! + ×^5/5! - x^7/7! + ...
cos x = 1 - ×^2/2! + x^4/4! - x^6/6! + ...
where x is (has to be) measured in radians.
@@cigmorfil4101p😊😊9
@cigmorfil4101 love ❤️ 😍 is pplp😅9l
@@cigmorfil41019p99p
@@cigmorfil4101😊
Is easy brother to solved
Bro, this is just a basic jee mains level question
Jee mains ka question hai
I love it
Олимпиадная зажача для класса 8-9? Сдишком уж лчевидное решение, можно было бы в лоб решать без замены, так как корни выдны сразу же. Обидно, что наше общество так сильно загнивает при капитализме...
тупо заголовок для привлечения, задачка более чем типовая
و لو كان معيار لديك اسمه بروتيكس أعطيه ذات الحصار بالحزاق مثلا x = 100 000
How would you write that 30/u after supposing 😂😂😂fault 😂😂😂
why does this feels like easy to asians
حتى ان اعطيته يا كوشنر دات المعيار ليس أساسا لان العد الاصلي و الحرفي ليس له أساسا اصلا .،
هده المعادلة مجنسة .، و اشم فيها عطر الحزاق متلا
В России это обычное задание примерно за 9 класс средней школы. На дом много таких задавали.
That's why math difficult
Just write x=±(pi/3) + pi×n
حتى مالانهاية الى نهاية واقفة .، فقط أعطه دات المعيار .،
X= 1 or x=2 . Max
You can solve this problem by taking natural logs on both sides. Much shorter route.
I think that you are thinking that log(A+B) = Log(A) + Log(B). That is not true. Rather Log(AB)=Log(A)Log(B) if A and B are both positive.
Well, log to the base 81, not natural logs, but same shit
@@irvingrabin I think he means taking log to the base 81 of each side, giving sin^2 (x) + cos^2 (x) = log,81(30), use the double angle formula of cos to create an expression in terms of sin, let u = sin(x), solve for u, substitute and find your answer that way
@@ow3nthehorror Incorrect usage of logarithmic rules aside, that method of finding x wouldn't even work considering you're essentially writing the equation as 1 = log81(30), meaning there's no x that'd satisfy it no matter what.
@@ennpii8147 sorry I’m just trying to learn here, what would be the correct way to apply the logarithm rules to this equation? Or is it not possible?
Just write both cos^2x and sin^2x in terms of cos2x. It will be way too easy 🛌
هده هي المعادلة تربط الاداتين المعياري الدي عير بمعيار لا علاقة له بزمان او مكان .، يرتبطان المعايرين بذات واهية لا وجود لها اصلا .، نحدد الاكس بدات المعيار من كوسينهات و السينيسهات فقط .،
I find that x=60 degree
لو كانت دات معيار لما كانت 30 تساوي مثلا
Very good
🙏
👍👍👍👍👍
معادلة دات 3 اقطار
8مليوم و 100 الف من إكس ؤس واحد + 8 تريليون و 100 مليار اكس ؤس 2 أو ثلاثة .
You're funny 😂
@@siriusman6169 سمتايز اكون عاريا
You overcomplicated this poor little thing. Here is how to destroy it.
First of all I'm introducing y=sin²x and noticing that cos²x=1-sin²x our equation becomes:
81^y+81^(1-y)=30
Not very beautiful. Let's divide by 9...
81^[y-1/2]+81^[1/2-y]=10/3.
Much better! Let's introduce z=81^[y-1/2]. Our equation becomes:
z+1/z=10/3.
We can see an obviouss solution, which is 3: 10/3=3+1/3.
And since z and 1/z are symetrical that means that 1/3 is also a solution.
But if we multiply our equation by z, we obtain a quadratic equation and we know that we have at most two solutions. We found two solutions, we're done.
So now we have to solve 81^[y-1/2]=3 or 81^[y-1/2]=1/3
Let's rewrite this as powers of 3: 3^[4(y-1/2)]=3 or 3^[4(y-1/2)]=3^(-1)
Then 4y-2=1 or 4y-2=-1
Then y=3/4 or y=-1/4.
But y=sin²x which means that the possible values of sin(x) are sqrt(3)/2, -sqrt(3)/2, 1/2 or -1/2
Then the possible values of x are pi/6, pi/3, 2.pi/3, 5.pi/6 and their opposites.
Can someone give me the 'quick' method which everyone is talking about in the comments?
hmm I don't know what 'quick' method everyone else is saying. but I did have my own method which was quicker than this. you begin by re-writing 81 as 3^4 and then multiplying the exponentials of both terms. Next, I realised that I could rewrite 30 as 3^3 + 3^1. This meant that I could equate the exponential so that when 4Sin^2x = 3, 4Cos^2x = 1. Then I multiplied the cosine function by 3 to get 12Cos^2x = 3 and then equated the sin and cosine function. Rearranging you would get that tan^2x = 3 meaning tanx = root3. Knowing your exact values, you should be able to tell that x = + or - 60 degrees otherwise +- pi/3 radians. Now i look back at the original equation and notice that sin(pi - x) = sin(x) and so 2pi/3 is also an answer. Next to find other solutions is when 4Sin^2x = 1 then 4cos^2x = 3. using the same method, you get tanx = 1/root3 which means x = 30 degrees or +-pi/6. from the solutions of +-pi/3 and +-pi/6 and 2pi/3 is also an answer you find that for each base angle of pi/3 or pi/6, there are solutions which repeat every pi/2.
I did it this way th-cam.com/video/j-sqIkjaLpY/w-d-xo.html
everyone seems to be talking about using log on both sides. But that method doesn't work.
it is easy
It takes too much long ago
Me being an IITian, who knows this is one of the most basic questions in JEE: 🥹
7:20
And sous❤si standard