Hello. Isn't simplier formula x=k*pi/2 where k is integer? Because even multipliers of pi are solution as well as odd ones so all multipliers of pi are solution but any multiplier of pi is an even multiplier of pi/2. and odd multipliers of pi/2 are also solution so all multipliers of pi/2 are solutions
That's exactly what I was thinking. Of course, having the three formulae for x shows how they've come from different branches, but I was half expecting you would wind up the video by showing how the three solution forms you've found can be simplified by merging them into one.
(sin x)^2 = (1 - cos 2x) /2 (cos x)^2 = (1 + cos 2x) /2 and then let y = 2 ^ (cos 2x) we can create a quadaric equation of y. final answer will be same.
Bro , we can also find imaginary solution by using Euler formula . Substituting cos^2(x) = e^-ix+e^ix /4 and similarly sin^2(x) = -1/4 (e^-ix - e^ix ) and get numerical solution in negative integer .
On aperçoit une racine évidente x = 0 et en fait x = n.pi va convenir; est-ce la seule infinité de solutions ? la réponse est oui pour cela on écrit l'équation (rac2)^(1-cos2x) + (rac2)^(1+cos2x) = 3 soit (rac2)^(-cos2x) + (rac2)^cos2x = 3/rac(2) soit encore ch[(cos2x)ln(rac2)] = (3/4)rac(2) et tu inverses le cosinus hyperbolique soit : cos(2x).ln(rac(2))= ln(rac(2)) soit cos(2x) = 1 et donc x = n.pi
A different method (looks simpler): Assume A=2^(sin²x), B=2^(cos²x); we can immediately get A*B = 2^(sin²x+cos²x)= 2; the original equation is A+B=3; from these two equations we can get (A, B) = (1, 2) or (2, 1); from here we can get sin²x = 1 or sin²x = 0; the rest is the same.
The odd multiples of pi together with the even multiples of pi gives the multiples of pi which are the even multiples of pi/2. Together with the odd multiples of pi/2 from the first case, we see that the solutions are all (integer) multiples of pi/2.
Writing y for sin^2(x) one gets 2^y+ 2/(2^y) = 3 2^(2*y) - 3 *(2^y) + 2 = 0 i.e ( 2^y -2)(2^y -1) = 0 Hereby sin^2(x) = y = 1, 0 Hereby angle x refers to all possible angles lying on the y axis and x axis
then α and β are the roots of u² -3u +2 = 0 (u-1)(u-2) = 0 u = 1 or 2 (2^sin²x, 2^cos²x) = (1,2) or (2,1) (sin²x, cos²x) = (0,1) or (1,0) (sinx, cosx) = (0,±1) or (±1,0) x = n π or n π /2, n ∈ Z x = k π /2, k ∈ Z
What if, instead of the equation being equal to a positive natural number, what if it equalled a fraction or an imaginary number, what would the solution be like? Would it be the same four answers only rotated, or would they be "squished" somehow?
I admire you so much bro ! I thought of the same approach but I rather multiplied both the sides by 2^cos²(x) to generate sin²(x) + cos²(x). Thankyou for the problem.
My solution, assuming x is real: Using everyone's favorite trig identity, the problem can be rewritten as 2^(cos^2(x)) + 2^(1-cos^2x)) = 3. Consider the function f(x) = 2^x + 2^(1-x). Then this function is differentiable everywhere, and f'(x) = ln2*(2^x - 2^(1-x)). From here it's easy to see that f'(x) < 0 for x
personally, i like to multiply the whole equation by 2^(cos^2 x ) to gete : 2^( 2 cos^2 x ) + 2 = 3 * 2^(cos^2 x ) so it becomes 2^( 2 cos^2 x ) - 3*2^( cos^2 x ) + 2 = 0
My thoughts on trying to solve this using algebra before watching the video: Let a = 2 ^ (sin^2 x), and b = 2 ^ (cos^2 x) So the question is a system of equations: a + b = 3 and a*b = 2 (because when you multiply things with the same base you add the power and sin^2 x + cos^2 x = 1 so 2^1 =2 Therefore, either 2 ^ (sin^2 x) = 2 and 2 ^ (cos^2 x) = 1 or the other way around. So we have 2 cases: Case 1: 2 ^ (sin^2 x) = 2 this means sin^2 x = 1 so (x = pi/2 + pi*k) (k = every number because the graph repeats every pi due to sin^2 instead of just regular sin) Case 2: 2 ^ (sin^2 x) = 1 this means sin^2 x = 0 (because 2^0 = 1) so (x = 0 + pi*k) So x = pi*k and x = pi/2 + pi*k Also pls don't roast me if I made a mistake anywhere, I'm only in 8th grade and trig is hard for me :)
Bro , we can also find imaginary solution by using Euler formula . Substituting cos^2(x) = e^-ix+e^ix /4 and similarly sin^2(x) = -1/4 (e^-ix - e^ix ) and get numerical solution in negative integer .
Thanks. I solved it while I was brushing my teeth. I took a pen to write the solution after. Good exercise for the brain. I like your channel for that. 👍
Why did you not combine the 3 results into one simple expression. In terms of PI/2, the 1st result being multiplied by (2n-1) gives all odd multiples of the term. For cos(x) = 1, extracting the term, we are left with (2k)*2 (multiplied by 2 to introduce the PI/2 term), or (4k), which give all even multiples divisible by 4. For cos(x) = -1, extracting the term, we are left with (2m-1)*2, or (4m-2), which gives all the other even multiples. The 2nd and 3rd ones combine to give all even multiples of PI/2 and the first gives all odd multiples of PI/2; so in total we have ALL multiples of PI/2. Therefore the final result is x = c*(PI/2) where c is any integer. Oh, and yes -- I know that with cos(x) being 1, 0 or -1, that encompasses all multiples of PI/2. I just wanted the logic of the results to show it also.
Sir, I have request for you to solve one problem.as per pythagorus theorem hypotenuse of right angle triangle is c^2 =a^2+b^2. This is same as 7×a/8 +b/2 fir any right angle triangle.can you help me solve this? Either by geometry,trigonometry or calculus?
Hey thanks for everything but I'd like to ask please about the way u found all the possible solutions by the means of *(2n-1)* I didn't really get this!
You can summarize the result with x=kπ/2
yes
Hello. Isn't simplier formula x=k*pi/2 where k is integer? Because even multipliers of pi are solution as well as odd ones so all multipliers of pi are solution but any multiplier of pi is an even multiplier of pi/2. and odd multipliers of pi/2 are also solution so all multipliers of pi/2 are solutions
yes
That's exactly what I was thinking. Of course, having the three formulae for x shows how they've come from different branches, but I was half expecting you would wind up the video by showing how the three solution forms you've found can be simplified by merging them into one.
th-cam.com/video/dkfa5L2qfak/w-d-xo.html
can you compare log2017(2018) and log2016(2017) please
combining all case we get multiple of pi/2
Nice!
(sin x)^2 = (1 - cos 2x) /2
(cos x)^2 = (1 + cos 2x) /2
and then
let y = 2 ^ (cos 2x)
we can create a quadaric equation of y.
final answer will be same.
An obvious solutions are when (sin(x) = 0, cos(x) = 1) or (sin(x) = 1, cos(x) = 0)
Let t= 2^sin(x)^2 ---> 2^cos(x)^2 = 2^(1-sin(x)^2 ) = 2/ t ---> t + 2/t = 3 ---> t^2 -3t + 2 = 0 ----> t= 1 and t= 2 etc ...
Bro , we can also find imaginary solution by using Euler formula . Substituting cos^2(x) = e^-ix+e^ix /4 and similarly sin^2(x) = -1/4 (e^-ix - e^ix ) and get numerical solution in negative integer .
On aperçoit une racine évidente x = 0 et en fait x = n.pi va convenir; est-ce la seule infinité de solutions ? la réponse est oui
pour cela on écrit l'équation (rac2)^(1-cos2x) + (rac2)^(1+cos2x) = 3 soit (rac2)^(-cos2x) + (rac2)^cos2x = 3/rac(2)
soit encore ch[(cos2x)ln(rac2)] = (3/4)rac(2) et tu inverses le cosinus hyperbolique soit :
cos(2x).ln(rac(2))= ln(rac(2)) soit cos(2x) = 1 et donc x = n.pi
A different method (looks simpler): Assume A=2^(sin²x), B=2^(cos²x); we can immediately get A*B = 2^(sin²x+cos²x)= 2; the original equation is A+B=3; from these two equations we can get (A, B) = (1, 2) or (2, 1); from here we can get sin²x = 1 or sin²x = 0; the rest is the same.
brilliant
great idea!!!
X equal npi
Unable to digest your argument in some of your problems ,
If we take cos²x = 1-sin²x and substitute y = sin²x , do we get the same solution??
BTW awesome explanation n presentation 👏 👌 👍🏻
Original Equation:
2↑sin²x + 2↑cos²x = 3
Substituting cos²x = 1 - sin²x:
2↑sin²x + 2↑(1 - sin²x) = 3
Substituting sin²x = y:
2↑y + 2↑(1 - y) = 3
2↑y + 2/2↑y = 3
2↑2y + 2 = 3·2↑y
2↑2y - 3·2↑y + 2 = 0
Rewriting exponent:
(2↑y)↑2 - 3·2↑y + 2 = 0
Factoring:
(2↑y - 1)(2↑y - 2) = 0
Here we get two solution paths:
2↑y = 1 (or) 2↑y = 2
Taking the binary log of both sides:
y = 0 (or) y = 1
Substituting y = sin²x
sin²x = 0 (or) sin²x = 1
sin²x = 0 → sinx = 0 → x = arcsin(0) = 2nπ, where n ∈ Ζ
sin²x = 1 → sinx = ±1 → x = arcsin(±1) = (2m + 1)π/2, where m ∈ Ζ
Combining solutions, x = kπ/2, where k ∈ Ζ
So yes, you will get the same solutions.
The odd multiples of pi together with the even multiples of pi gives the multiples of pi which are the even multiples of pi/2. Together with the odd multiples of pi/2 from the first case, we see that the solutions are all (integer) multiples of pi/2.
Writing y for sin^2(x) one gets
2^y+ 2/(2^y) = 3
2^(2*y) - 3 *(2^y) + 2 = 0
i.e ( 2^y -2)(2^y -1) = 0
Hereby sin^2(x) = y = 1, 0
Hereby angle x refers to all possible angles lying on the y axis and x axis
α+β=3, αβ=2
Good thinking!!! 🤩
That's also how I did it. Very nice.
then α and β are the roots of u² -3u +2 = 0
(u-1)(u-2) = 0
u = 1 or 2
(2^sin²x, 2^cos²x) = (1,2) or (2,1)
(sin²x, cos²x) = (0,1) or (1,0)
(sinx, cosx) = (0,±1) or (±1,0)
x = n π or n π /2, n ∈ Z
x = k π /2, k ∈ Z
Simple question based on fundamentals of quadratic equation and trigonometry.
If cos^2x=0 or 1 then directly x=kπ/2
Thanks a lot 🌹🌹🇮🇳
Np. Thank you
X=0 or X= pi/2 [2pi]
like bruh isn't that too easy??
Pi/2
جيد ❤️❤️
លំហាត់ល្អ!👍👍
សូមអរគុណ!
Wow...U can write Khmer too!
My first thought🤦🏻♀️: ln2^sin²x+ln2^cos²x=ln3.....sin²xln2+cos²xln2=ln3....ln2(sin²x+cos²x)=ln3....ln2=ln3?!?!?😆
Seriously? 😜
Answer could just be x=k/2 pi
Arey bhai, you should write what to find on the screen. I am trying to proove this equation 🤣
😁
x = n*(π/2).
Le non du logiciel 🙏
Notability
Not understand
School books problem
I don't. Understand. 2Kп
integer multiples of 2pi bring us to the same point on the unit circle
Non sense question
sin^2(x) = 0 => x = Pi*k
sin^2(x) = 1 => x = Pi/2 + Pi*k
If we combine these roots we can write x = k * (Pi/2) where k - integer.
Yes Sir. I am also a mathematics teacher and I did enjoy & appreciate your explanation. Thanks a million.
Absolutely! Thank you for the feedback!!! 🥰
What if, instead of the equation being equal to a positive natural number, what if it equalled a fraction or an imaginary number, what would the solution be like? Would it be the same four answers only rotated, or would they be "squished" somehow?
I admire you so much bro !
I thought of the same approach but I rather multiplied both the sides by 2^cos²(x) to generate sin²(x) + cos²(x). Thankyou for the problem.
My solution, assuming x is real:
Using everyone's favorite trig identity, the problem can be rewritten as 2^(cos^2(x)) + 2^(1-cos^2x)) = 3.
Consider the function f(x) = 2^x + 2^(1-x). Then this function is differentiable everywhere, and f'(x) = ln2*(2^x - 2^(1-x)). From here it's easy to see that f'(x) < 0 for x
Nice!
personally, i like to multiply the whole equation by 2^(cos^2 x ) to gete : 2^( 2 cos^2 x ) + 2 = 3 * 2^(cos^2 x )
so it becomes 2^( 2 cos^2 x ) - 3*2^( cos^2 x ) + 2 = 0
Amazing!!! A lot of different concepts united together!
Glad you think so!
My thoughts on trying to solve this using algebra before watching the video:
Let a = 2 ^ (sin^2 x), and b = 2 ^ (cos^2 x)
So the question is a system of equations: a + b = 3 and a*b = 2 (because when you multiply things with the same base you add the power and sin^2 x + cos^2 x = 1 so 2^1 =2
Therefore, either 2 ^ (sin^2 x) = 2 and 2 ^ (cos^2 x) = 1 or the other way around. So we have 2 cases:
Case 1:
2 ^ (sin^2 x) = 2
this means sin^2 x = 1 so (x = pi/2 + pi*k) (k = every number because the graph repeats every pi due to sin^2 instead of just regular sin)
Case 2:
2 ^ (sin^2 x) = 1
this means sin^2 x = 0 (because 2^0 = 1) so (x = 0 + pi*k)
So x = pi*k
and x = pi/2 + pi*k
Also pls don't roast me if I made a mistake anywhere, I'm only in 8th grade and trig is hard for me :)
So actually, the final solution is n*pi/2, because cos(x) is equal to -1,0 or 1 every pi/2.
Yessir! 😁
x=0degree
so 2^0+2^1=3
How x=(2n - 1)π/2
I took the shortcut to get to the end and split 3 into 2^1 + 2^0. So either sin^2(x) = 1 and cos^2(x) = 0 OR sin^2(x) = 0 and cos^2(x) = 1.
It's so easy that I solved it in mind while seeing the thumbnail love from india 🇮🇳🇮🇳
A normal que for jee aspirant.
hmm
x = pi/2*k, k is an integer
* what if we difine 2 to the power of something can somereason to be zero and branch new to "real Virtual Number" ? 17th Febr 0002022 *
solution set of cos² x = 1 is
I. {1, -1}
II. {(1,0), (-1,0)}
III. { O, π }
A. I and II
B. II and III
C. I , II, III
D. None of them
Its easy sin aur cos k max value -1to1 hoti hai aur square lga hai to 0-1 sum 3 hona chahiye to kisi ek ka exponent 0 hoga aur ek ka 1 so x=n(π/2)
Nice question.
I was lazy.
2^(sinx)^2(1+2^(1-2(sinx)^2)=3
When sinx=1 and sinx=0 the equation satisfied and hence the answer.
It’s good to be lazy sometimes! 😉
(cosx)^2=1 => (sinx)^2=0, then you don't have to deal with 2 cases.
Which device do you use to write ....(ipad or lcd board ....or tablet ...or ??????
K(π/2) , k€ Z ...is the solution...since it satisfies the given equation...
If we put log(2,z)=log of z in base 2, we get log(2,3)=1 !!!!!
A good not difficult problem but not IMO material.
Phương trình gọn gàng.
If you substitute sin2x with a you get quadratic
🙏🏼 good afternoon, answer sharing x=90°
Put x equal to 45^0 the statement reduces to 2√2 = 3 🤣
The last case is just x=kpi, k=0, 1, -1, 2, -2,...
Easily sol. Thita =0°{let}😊😊😊
Bro , we can also find imaginary solution by using Euler formula . Substituting cos^2(x) = e^-ix+e^ix /4 and similarly sin^2(x) = -1/4 (e^-ix - e^ix ) and get numerical solution in negative integer .
This would work, except that those are the wrong formulas
@@sweetcornwhiskey what you mean to say , you didn't know about Euler sinx and cos x formulas
i think similar kind of ques ask in jee main exam
0, pi/2, pi, 3pi/2,... just looking
Thanks. I solved it while I was brushing my teeth. I took a pen to write the solution after. Good exercise for the brain. I like your channel for that. 👍
YOU are AMAZING! 😍
We're lucky you weren't in the toilet !
@@WahranRai 😂
crack
Put sine x equals cos 90 - x. Ln e.
The x values are 0,90,180
x = k*pi/2 where k is an integer
This is sleight of hand.
What kind of math is this?
I can solve it easley
Sir ji what is voi ???
Why did you not combine the 3 results into one simple expression. In terms of PI/2, the 1st result being multiplied by (2n-1) gives all odd multiples of the term. For cos(x) = 1, extracting the term, we are left with (2k)*2 (multiplied by 2 to introduce the PI/2 term), or (4k), which give all even multiples divisible by 4. For cos(x) = -1, extracting the term, we are left with (2m-1)*2, or (4m-2), which gives all the other even multiples. The 2nd and 3rd ones combine to give all even multiples of PI/2 and the first gives all odd multiples of PI/2; so in total we have ALL multiples of PI/2. Therefore the final result is x = c*(PI/2) where c is any integer.
Oh, and yes -- I know that with cos(x) being 1, 0 or -1, that encompasses all multiples of PI/2. I just wanted the logic of the results to show it also.
I did this way
2^sin^2x+2^cos^2x=2^1+2^0
Now equate exponents by comparing 2^sin^2x=2^1=> x=π/2 for all x belonging to [0,π/2]
Sir, I have request for you to solve one problem.as per pythagorus theorem hypotenuse of right angle triangle is c^2 =a^2+b^2. This is same as 7×a/8 +b/2 fir any right angle triangle.can you help me solve this? Either by geometry,trigonometry or calculus?
2n+1
Y=0
Cos square x=0
Cos x=0
Cos x=cos 0
X=0
Substitute in eq 1 in lhs
2+1=3=rhs
Hence proved
1 mo ago
Nice questions 👌👌👌 mast
Thanks
The answer should be further simplified to k*pi/2. This expression has compactly integrated the 3 solutions you presented.
Eh?💀
The equation was easy to solve. The solutions will be into the form k*90° or k*(1.5708 rad) where k is any relative integer.
X=0
Hey thanks for everything but I'd like to ask please about the way u found all the possible solutions by the means of *(2n-1)* I didn't really get this!
어떤그리움^^
One could maybe also use 1-sin(x)^2 rather than 1-cos(x)^2...
And give "One answer" as x=n*π/2...
sin30.=1/2
Took me awhile but I did it myself.
And at this point I don’t know to be proud or disappointed.
a very similar question appeared in the jee mains exam
Did it? Good to hear!
x = n pi / 2, where n is an integer (negative, zero, or positive all work) is the entire solution.
x=0,,x=90 deg.,x=270deg.ans
O bucurie sa vezi ASTFEL de probleme rezolvare elegant! Mai vrem!
cos^2(x)=1 has the same solutions as sin(x)=0 (x=kpi)
Excelente
Thanks!
Nice - This is the most disguised quadratic I have seen!
Essentially this is a quadratic equation.
X can be 0 and pi/2 in for x belongs to ( 0,pi )
greetings from Turkey
Greetings from the United States
I am from Bangladesh
Hi there! Greetings from the United States! 😍
Can we use other version of answer like if the cosx = 0, x = 2πk + π/2 where k is an integer ?
You got it! 🤩
Marvellous. Thank you.
It's so long method
Why?