Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Solve the equation: $\sin^25x+\sin^23x = 1+\cos(8x)$.

I tried : $1+\cos(8x) = 2\cos^2(4x)$ which gives :

$$\begin{align*} \sin^25x+\sin^23x &= 2\cos^2(4x)\\ &= 2(1-\sin^2(4x))\\ &= 2-2\sin^2(4x)\\ \end{align*}$$

share|improve this question

2 Answers 2

$$2\sin^25x-1+2\sin^23x-1=2\cos8x$$ $$-\cos10x-\cos6x=2\cos8x$$ $$-2\cos8x\cos2x=2\cos8x$$ Should be easy from there.

share|improve this answer

As $\cos(A-B)\cos(A+B)=\cos^2A-\sin^2B$

$\cos8x=\sin^25x+\sin^23x-1=-(\cos^23x-\sin^25x)=-\cos(5x-3x)\cos(5x+3x)=-\cos8x\cos2x$

$\cos8x(1+\cos2x)=0$

If $\cos8x=0,8x=(2n+1)\frac\pi2,x=(2n+1)\frac\pi{16}$

If $1+\cos2x=0\implies \cos2x=-1=\cos\pi,2x=(2m+1)\pi,x=(2m+1)\frac\pi2$ where $m,n$ are any integer

share|improve this answer
    
thanks a lot... –  Sachin Sharmaa Feb 18 '13 at 7:06
    
@SachinSharmaa, my pleasure. But, the solution won't have been so easy if the right hand side did not contain $\cos8x$ as a factor. –  lab bhattacharjee Feb 18 '13 at 15:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.