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

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

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

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

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



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

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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

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