Assign sinx = 0, ±1 on the complex plane, and every additional right angle starting from 0 will fulfill one of the values. Therefore we can simplify the solution as "x = ±kπ/2"
If the answers are sinx=0 OR sinx=1 OR sinx=-1, then any x=½kπ will correspond to the correct answer. There is no need for four different solutions. Lastly, thank you, your explanation is great.
@oahuhawaii2141 this step they always do in algebra but in geometry it is so much easier. Completely agree. Also, somewhere halfway he did a complicated two step process to show that u*u = u^2. 😂
I love your solution, but why choose "x" for your substitution variable at 3:55? Using the same variable there could be confusing for some students. Highly recommend you choose a different variable for your substitutions.
Hi every One. I do appreciate the way you used to solve the Equation. However, I faced a difficulty when I was trying to solve the following equation: 7^x + 1 = 8^x. Can give me a hand on that, please?
Hi luis with my little knowledge I think the best way to solve this is to consider the monotonicity of functions So 7^x +1=8^x (7/8)^x +(1/8)^x=1 now here we have two exponential function on the left and both is monotonically decreasing over all real numbers so the sum of the two monotonically decreasing function is always decreasing in the same interval (here all real numbers) But the right hand side of the equation is a constant Thus the LHS and RHS function will intersect at only one point meaning there is only one solution exist for the equation And in this case it is easy to notice that the solution is x=1 Hence the answer to the equation is x=1 Hope you will be able to understand my explanation
BTW, at 03:57 in your video, you made a variable for 7^sin²(x) , but you chose to use x instead of y . This is an illegal step in your math logic, since you must be able to distinguish between the meaning of 'x' -- either the original unknown x or the newly made x to substitute for a complicated term utilizing x . Bad: x = 7^sin²(x) . Good: y = 7^sin²(x) .
@@priyankachhajer1702: The way they're written makes them hard to distinguish on my small screen. If she wrote the uppercase 'X' to appear distinctly different from lowercase 'x', then it may be OK. I'm sure a student at the back of the class may get confused.
Put x equals to zero which satisfies the equation So, the answer is x equals to zero.This is only particular solution. Similarly, pi over 2 is also a particular solution.
I thought the unknowns were sine x and cosin x. I solved this in a different way and got the same answer and faster, but I still like your method. I think you should have mentioned what is the unknown and also once you find that sin (x) =+/-1, 0, you should use the first trig identity to find the value of the cosine^2(x). The whole idea was to apply algebra in a quick way to arrive to he conclusion. Keep up the good work.
Hopefully you would use ,(comma) carefully. Comma means and, so x-1=0,x-7=0 is not proper expression. X-1=0 or x-7=0 would be correct. Mathematics is always related with logics, so 'and' and 'or' must be used correctly.
Eu podia substituir 7 elevado a cos X por y e daí eu resolvo a equação do 2 grau e depois substituir os valores no x da equação característica para ver se realmente confere com a igualdade.
I think I left out some of the solutions where sin(x) = -1. That means x = -pi/2+2ki. So....the general solution is actually x = kpi/2...every single integer multiple of pi/2.
Simply use hit and trial method try putting in 0. U will get your 1st ans. Now just use the same 0,1 , 1,0 values from the unit circle you will get all your answers 😊
Легче возвести обе стороны в квадрат... Дальше спецы поймут. Просто я не могу применять математические знаки в чате, чтобы описать это. Приводится к квадратному уравнение а2 -59а+49=0
That's just extra work that accomplishes nothing. (And in your extra work, you made an error!) I can prove your method does nothing useful. If you let B represent the base 7, and notice that the RHS constant 8 is B + 1 , the original equation is: B^cos²(x) + B^sin²(x) = B + 1 Squaring both sides yields: (B^cos²(x) + B^sin²(x))² = (B + 1)² (B^cos²(x))² + 2*(B^cos²(x))*(B^sin²(x)) + (B^sin²(x))² = B² + 2*B + 1 (B²)^cos²(x) + 2*B^(cos²(x)+sin²(x)) + (B²)^sin²(x) = B² + 2*B + 1 (B²)^cos²(x) + 2*B^(1) + (B²)^sin²(x) = B² + 2*B + 1 (B²)^cos²(x) + (B²)^sin²(x) = (B²) + 1 Compare this with the original equation: B^cos²(x) + B^sin²(x) = B + 1 They're very similar. Each iteration of squaring both sides of the equation simply squares (B) in the formula. And another iteration yields: (B⁴)^cos²(x) + (B⁴)^sin²(x) = (B⁴) + 1 The general formula for squaring is: (Bᵐ)^cos²(x) + (Bᵐ)^sin²(x) = (Bᵐ) + 1 with m = 2ⁿ, and n = 0, 1, 2, 3, ... is the iteration number. The original equation has n = 0 . Note that n can be negative integers, too. For n = -1 , we have: (√B)^cos²(x) + (√B)^sin²(x) = (√B) + 1 Anyway, an easy solution to the original general equation uses the trigonometric identity: cos²(x) + sin²(x) = 1 B^cos²(x) + B^sin²(x) = B + 1 B^cos²(x) + B^(1 - cos²(x)) = B + 1 B^cos²(x) + B^1 / B^cos²(x) = B + 1 B^cos²(x) - (B + 1) + B / B^cos²(x) = 0 [B^cos²(x)]² - (B + 1)*[B^cos²(x)] + B = 0 [B^cos²(x) - 1]*[B^cos²(x) - B] = 0 B^cos²(x) = 1, B cos²(x) = log(1)/log(B), log(B)/log(B) = 0, 1 cos(x) = 0, ±1 x = ±π/2 + 2*π*k, 2*π*k, π + 2*π*k, k ∈ ℤ x = ±90⁰ + k*360⁰, 0⁰ + k*360⁰, 180⁰ + k*360⁰ That means all integer multiples of 90⁰ or π/2 are solutions: x = n*π/2, n ∈ ℤ Note that the solutions are independent of B .
Yes, he messed up with his math: Bad: "x" = 7^sin²(x) . Good: y = 7^sin²(x) . Also, he repeated -π/2 as 3*π/2 , and missed π/2 . For a complete _long_ answer, start from: sin(x) = 0, ±1 For sin(x) = ±1: x = ..., ±π/2 - 2*π, ±π/2, ±π/2 + 2*π, ... = π/2*{ ..., ±1 - 4, ±1, ±1 + 4 , ... } = π/2*{ ..., -5, -3, -1, 1, 3, 5, ... } x = π/2*{ odd integers } For sin(x) = 0: x = ..., -2*π, -π, 0, π, 2*π, 3*π, ... = π*{ ..., -2, -1, 0, 1, 2, 3, ... } = π/2*{ ..., -4, -2, 0, 2, 4, 6, ... } x = π/2*{ even integers } Combine x solutions together as: x = π/2*{ odd integers } or π/2*{ even integers } = π/2*{ odd or even integers } = π/2*{ all integers } x = n*π/2 , n ∈ ℤ An easier way is to list the 4 parts of x in degrees and sort the order: x = 0⁰ + k*360⁰, 180⁰ + k*360⁰, ±90⁰ + k*360⁰ x - k*360⁰ = 0⁰, 180⁰, 90⁰, -90⁰ x - k*360⁰ = -90⁰, 0⁰, 90⁰, 180⁰ That's all multiples of 90⁰ or π/2: x = n*π/2 , n ∈ ℤ
Great, the solution was very nice
x={0;90}in deg.
Your step by step explanation portrays the beauty of Maths. Excellent job.
It's so simple
Sin²x = 1-Cos²x
Make quadratic and solve 👍
Assign sinx = 0, ±1 on the complex plane, and every additional right angle starting from 0 will fulfill one of the values. Therefore we can simplify the solution as "x = ±kπ/2"
Yeah I just saw that.
If the answers are sinx=0 OR sinx=1 OR sinx=-1, then any x=½kπ will correspond to the correct answer. There is no need for four different solutions. Lastly, thank you, your explanation is great.
Thank you.
Just mark the points on the unit circle that correspond to the 4 angles.
@oahuhawaii2141 this step they always do in algebra but in geometry it is so much easier. Completely agree.
Also, somewhere halfway he did a complicated two step process to show that u*u = u^2. 😂
Thank you for making the solution look so simple.
The problem looks good 😊😊
I love your solution, but why choose "x" for your substitution variable at 3:55? Using the same variable there could be confusing for some students. Highly recommend you choose a different variable for your substitutions.
Nice video
Anybody attempting olympiad problem should be able to proceed faster. This is excruciatingly slow.
I almost solved it ❤❤❤❤❤
Nice solution but music is loud and distracting.
Hi every One. I do appreciate the way you used to solve the Equation. However, I faced a difficulty when I was trying to solve the following equation: 7^x + 1 = 8^x. Can give me a hand on that, please?
Hi luis
with my little knowledge I think the best way to solve this is to consider the monotonicity of functions
So
7^x +1=8^x
(7/8)^x +(1/8)^x=1
now here we have two exponential function on the left and both is monotonically decreasing over all real numbers so the sum of the two monotonically decreasing function is always decreasing in the same interval (here all real numbers)
But the right hand side of the equation is a constant
Thus the LHS and RHS function will intersect at only one point meaning there is only one solution exist for the equation
And in this case it is easy to notice that the solution is x=1
Hence the answer to the equation is x=1
Hope you will be able to understand my explanation
The sol is so simple but he is doing complicated
1sec q for jee asp
7^cos²(x) + 7^sin²(x) = 8 { Use sin²(x) = 1 - cos²(x) }
7^cos²(x) + 7^(1 - cos²(x)) = 8
Let y = 7^cos²(x) ---> cos(x) = ±√log₇(y)
y + 7/y = 8
y² - 8*y + 7 = 0
(y - 1)*(y - 7) = 0
y = 1, 7
cos(x) = ±√log₇(1), ±√log₇(7)
= ±√0, ±√1
cos(x) = 0, ±1
x = k*π/2, k integer
BTW, at 03:57 in your video, you made a variable for 7^sin²(x) , but you chose to use x instead of y . This is an illegal step in your math logic, since you must be able to distinguish between the meaning of 'x' -- either the original unknown x or the newly made x to substitute for a complicated term utilizing x .
Bad: x = 7^sin²(x) .
Good: y = 7^sin²(x) .
And at the end, you repeat -π/2 as 3*π/2 , and miss π/2 .
@@oahuhawaii2141 I think using capital X and small x is allowed as we use A,B,C as vertices of triangle and use a,b,c as sides of triangle…
@@priyankachhajer1702: The way they're written makes them hard to distinguish on my small screen. If she wrote the uppercase 'X' to appear distinctly different from lowercase 'x', then it may be OK. I'm sure a student at the back of the class may get confused.
We can directly write 7^cos²x as 7^1-sin²x and solve further it will be solved in just 4-5 lines
You forgot to put parentheses:
7^(1-sin²(x))
Simpler way is a = 7^cos^2x
and sin^2x = 1 - cos^2x
This gives a^2-8a+7=0
Put x equals to zero which satisfies the equation So, the answer is x equals to zero.This is only particular solution. Similarly, pi over 2 is also a particular solution.
Why to use so lengthy method ? Just break the 8 into (7¹+7⁰) and consider the cases. So simple!!!!!!!
X=0,pi/2,pi,3*pi/4
x = k pi / 2, (k is integer)
3pi/2 = - pi/2
I thought the unknowns were sine x and cosin x. I solved this in a different way and got the same answer and faster, but I still like your method. I think you should have mentioned what is the unknown and also once you find that sin (x) =+/-1, 0, you should use the first trig identity to find the value of the cosine^2(x). The whole idea was to apply algebra in a quick way to arrive to he conclusion. Keep up the good work.
Thanks for Appreciation.
It's very clear that 7 + 1 = 8
So x = 90 or 0
Multiple of π/2
One thing I like to say at time slot 9-35 u sroke out sin from both sides.
It is not a correct method.
U know sin without any angle is meaningless.
7^cosx^2+7^sinx^2= 8-1=7=8/1=8=7 tada sinx2+cosx^2=1]=[ sinx^2+cosx^2= 2 sinx+cosx= 1=1 sinx+cosx=1-1=1/1=1 sinx=O cosx=1 X=2pik pi=4 k=1 X=8;p
writing z for 7 ^ ( sin^2 ( x) ) one gets
7 ^ ( cos^2 ( x) )
= 7 / ( 7 ^ ( sin^2 ( x) ) )
= 7 / z
Hereby z + 7 / z = 8
z^2 - 8 z + 7 = 0
z = 4 + 3, 4 - 3
7 ^ ( sin^2 ( x)) = z = 7 ^ 1, 7 ^0
Hereby, sin^2 ( x) = 1, 0
sin( x) = - 1, 0 , 1
x = n π, ( n + 1/2) π
or x = m π /2 , for integral m
What if we multiply both sides by 7 raised to cos x squared? Are we going to have the same solutions?
😮
since -pi/2=3pi/2 {pi/2,pi,3pi/2,2kpi} ....> (every pi/2 round up) justifies
sin²x + cos²x= 1 🤔
1.sin²x=1, cos²x=0
sinx=1, sinx= -1, cos²x=0
2. sin²x=0, cos²x=1
sinx=0, cosx=1, cosx=-1
3. sin²x=0,5 cos²x=0,5
sinx=cosx
x= п/4 ,
п/4=45° ==> corner balance !!
7⁰⁵+7⁰⁵ = ?!
7^x=4 ?!
This answer is to long. We can take cos²x=1-sin²x
Приветствую автора. Не стоит применять одну и ту же переменную для замены. Это может привести к путанице.
Bad: x = 7^sin²(x) .
Good: y = 7^sin²(x) .
Hopefully you would use ,(comma) carefully. Comma means and, so x-1=0,x-7=0 is not proper expression. X-1=0 or x-7=0 would be correct. Mathematics is always related with logics, so 'and' and 'or' must be used correctly.
7^(1-sin²x) +7^sin²x=8
7÷7sin²x +7^sin²x=8
7 +7sin^4(x)=8.7sin²x
Let 7^sin²x=A
7+A²=8A
A²-8A+7=0
(A-7)(A-1)=0
A=7, A=1
sin²x=7 --> sinx=±√7 ❎
sin²x=1 --> sinx=±1
x=π/2, x=3π/2
Eu podia substituir 7 elevado a cos X por y e daí eu resolvo a equação do 2 grau e depois substituir os valores no x da equação característica para ver se realmente confere com a igualdade.
0 or 90°
My doubt is that can we do 7^sin2x + 7^cos2x = 7 + 1 = 7^1 + 7^0
Thus, sin2x + cos2x = 1+0 = 1
Thus sinx = ±1 or 0, and cosx = ±1 or 0
Alternately write cos in terms of Sine or sine in terms of cos and proceed
Yes and thus he is not able to do JEE advance maths😂🤣
To me, this is cleanest general solution ...
Let B = 7 .
B^cos²(x) + B^sin²(x) = B + 1
= B^1 + 1
= B^(cos²(x) + sin²(x)) + 1
= B^cos²(x) * B^sin²(x) + 1
0 = B^cos²(x) * B^sin²(x) - B^cos²(x) - B^sin²(x) + 1
0 = [ B^cos²(x) - 1 ] * [ B^sin²(x) - 1 ]
B^cos²(x) = 1 , B^sin²(x) = 1
cos²(x) = 0 , sin²(x) = 0
cos(x) = 0 , sin(x) = 0
x = π/2 + π*k , x = π*k , k ∈ ℤ
x = π/2*(1 + 2*k) , x = π/2*(2*k) , k ∈ ℤ
x = π/2*{odd integers} , x = π/2*{even integers}
x = π/2*{all integers}
x = n*π/2 , n ∈ ℤ
Note that the solutions are independent of B for: B^cos²(x) + B^sin²(x) = B + 1 .
What is the unit of x
Angle we take in degree.
0° or 90°
Background sound distract 😕😕 kar raha hhh
Okk will try to change.
7/7 sin^2 (x ) + 7 sin ^2 ( x) = 8
Sin^2 ( x) = 1
X = n pi ( n= 1/2 ,3/2 ,5/2 = ( m + 1)/2 & (m] any even integer [ ) ?
X=0 satisfy the equation…
I solved this in just 14 steps in less than 3 minutes
Unnecessarily u have increased the steps.
For beginners.
The solution is are 90 degree angles.
7^(sin^2(x))=1 or 7 and there are many solutions...x = kpi or x = pi/2+2kpi
I think I left out some of the solutions where sin(x) = -1. That means x = -pi/2+2ki. So....the general solution is actually x = kpi/2...every single integer multiple of pi/2.
X=90
7^{(sinx)^2}=7
(Sinx)^2=1
sinx=1
Sinx=sin90
X=90
On the other hand
7^{(sinx)^2=1
Sinx=sin0
X=0, (may be )
X = 90° & 0°
What about the cos²(x) part that you forgot?
General solution:
x = n*90⁰, n integer.
Unique solution:
x = 0⁰, ±90⁰, 180⁰
In solution list pi/2 is missing.
Yes, he repeated -π/2 with 3*π/2 .
Don"t rational!!
X = zero degree
Simply use hit and trial method try putting in 0. U will get your 1st ans. Now just use the same 0,1 , 1,0 values from the unit circle you will get all your answers 😊
Why Add a 2kπ?
Wrongly solved
The part where he writes x = 7^sin²(x) is wrong because the left-side x should be a different variable.
He also missed one fourth of the solutions by repeating -π/2 with 3*π/2 . He neglected π/2 .
7^0 = 1 not 0
Before watching:
7^0 = 1, 7^1 = 7,1+7=8.
Thus, the solutions occur at nπ/2 , where n is any integer.
What about the other solutions? You're just guessing that you've found all of them.
Minor mistake
?
Un comentario: sqr(1)=+1 solamente
Also -1.
@@haiderlughmani no. F(x)=sqr(x) siempre es >=0 en los números reales
п/2, п, …., n*п/2. 7^0+7^1=8
Легче возвести обе стороны в квадрат... Дальше спецы поймут. Просто я не могу применять математические знаки в чате, чтобы описать это. Приводится к квадратному уравнение а2 -59а+49=0
That's just extra work that accomplishes nothing. (And in your extra work, you made an error!) I can prove your method does nothing useful. If you let B represent the base 7, and notice that the RHS constant 8 is B + 1 , the original equation is:
B^cos²(x) + B^sin²(x) = B + 1
Squaring both sides yields:
(B^cos²(x) + B^sin²(x))² = (B + 1)²
(B^cos²(x))² + 2*(B^cos²(x))*(B^sin²(x)) + (B^sin²(x))² = B² + 2*B + 1
(B²)^cos²(x) + 2*B^(cos²(x)+sin²(x)) + (B²)^sin²(x) = B² + 2*B + 1
(B²)^cos²(x) + 2*B^(1) + (B²)^sin²(x) = B² + 2*B + 1
(B²)^cos²(x) + (B²)^sin²(x) = (B²) + 1
Compare this with the original equation:
B^cos²(x) + B^sin²(x) = B + 1
They're very similar. Each iteration of squaring both sides of the equation simply squares (B) in the formula. And another iteration yields:
(B⁴)^cos²(x) + (B⁴)^sin²(x) = (B⁴) + 1
The general formula for squaring is:
(Bᵐ)^cos²(x) + (Bᵐ)^sin²(x) = (Bᵐ) + 1 with m = 2ⁿ, and n = 0, 1, 2, 3, ... is the iteration number. The original equation has n = 0 . Note that n can be negative integers, too. For n = -1 , we have:
(√B)^cos²(x) + (√B)^sin²(x) = (√B) + 1
Anyway, an easy solution to the original general equation uses the trigonometric identity:
cos²(x) + sin²(x) = 1
B^cos²(x) + B^sin²(x) = B + 1
B^cos²(x) + B^(1 - cos²(x)) = B + 1
B^cos²(x) + B^1 / B^cos²(x) = B + 1
B^cos²(x) - (B + 1) + B / B^cos²(x) = 0
[B^cos²(x)]² - (B + 1)*[B^cos²(x)] + B = 0
[B^cos²(x) - 1]*[B^cos²(x) - B] = 0
B^cos²(x) = 1, B
cos²(x) = log(1)/log(B), log(B)/log(B) = 0, 1
cos(x) = 0, ±1
x = ±π/2 + 2*π*k, 2*π*k, π + 2*π*k, k ∈ ℤ
x = ±90⁰ + k*360⁰, 0⁰ + k*360⁰, 180⁰ + k*360⁰
That means all integer multiples of 90⁰ or π/2 are solutions:
x = n*π/2, n ∈ ℤ
Note that the solutions are independent of B .
Глупый ответ. Разве нельзя сделать объединение ответов?
И как можно делать замену переменной, используя искомую переменную?
Yes, he messed up with his math:
Bad: "x" = 7^sin²(x) .
Good: y = 7^sin²(x) .
Also, he repeated -π/2 as 3*π/2 , and missed π/2 .
For a complete _long_ answer, start from:
sin(x) = 0, ±1
For sin(x) = ±1:
x = ..., ±π/2 - 2*π, ±π/2, ±π/2 + 2*π, ...
= π/2*{ ..., ±1 - 4, ±1, ±1 + 4 , ... }
= π/2*{ ..., -5, -3, -1, 1, 3, 5, ... }
x = π/2*{ odd integers }
For sin(x) = 0:
x = ..., -2*π, -π, 0, π, 2*π, 3*π, ...
= π*{ ..., -2, -1, 0, 1, 2, 3, ... }
= π/2*{ ..., -4, -2, 0, 2, 4, 6, ... }
x = π/2*{ even integers }
Combine x solutions together as:
x = π/2*{ odd integers } or π/2*{ even integers }
= π/2*{ odd or even integers }
= π/2*{ all integers }
x = n*π/2 , n ∈ ℤ
An easier way is to list the 4 parts of x in degrees and sort the order:
x = 0⁰ + k*360⁰, 180⁰ + k*360⁰, ±90⁰ + k*360⁰
x - k*360⁰ = 0⁰, 180⁰, 90⁰, -90⁰
x - k*360⁰ = -90⁰, 0⁰, 90⁰, 180⁰
That's all multiples of 90⁰ or π/2:
x = n*π/2 , n ∈ ℤ
x = 90° or 0°
Leading to
7^0 + 7^1
Or
7^1 + 7^0
I did it orally
You're missing x = -90⁰, 180⁰ .