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Solve the following equation over the real numbers: $$ \sin(2x)\sin(x)+ \cos^2(x) = \sin(5x)\sin(4x)+ \cos^2(4x) $$

What I've tried so far, to no avail, is using identities (product-to-sum and sum-to-product, as well as the double angle identity) in search of a more convenient form, like a product equaling $0$ or a polynomial through some notation. Alternate forms I've reached : $$ \sin(5x)\sin(3x)=\sin(6x)\sin (3x)$$ $$\sin(x)(\sin(x)(2\cos(x)-1)- 4\cos(x)(1-2\sin^2(x))(\sin(5x)-\sin(4x) ))=0$$ Perhaps a helpful identity: $$ \cos^2(x)- \cos^2(4x)= \sin(5x)\sin(3x) $$

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  • $\begingroup$ use $cos^2(a)+sin^2(a) = 1$ first on both sides then factorize. Now use $sin(A)-sin(B)$ as a product on both sides and simplify (giving some roots), then write $sin(C)cos(D)$ as a difference of sines on both sides and simplify. finish with a difference of sines written as a product and solve. $\endgroup$ – Paul Apr 8 at 14:25
  • $\begingroup$ I've difficulties understanding your suggestion $\endgroup$ – Luca Pana Apr 8 at 14:29
  • $\begingroup$ Give the first step a go though to replace the $cos^2$ terms then see where it goes.. $\endgroup$ – Paul Apr 8 at 14:31
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Hint

By $\cos2y=2\cos^2y-1$ and

http://mathworld.wolfram.com/WernerFormulas.html and

http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html,

$$\cos x-\cos3x+1+\cos2x=\cos x-\cos9x+1+\cos8x$$

$$\cos2x-\cos8x=\cos3x-\cos9x$$

$$2\sin5x\sin3x=2\sin3x\sin6x$$

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  • $\begingroup$ yes this was an alternate form I reached, when posting I messed up a sign $\endgroup$ – Luca Pana Apr 8 at 14:38
  • $\begingroup$ @Luca, So we are done? $\endgroup$ – lab bhattacharjee Apr 8 at 14:39
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Use $\cos^2(a)+\sin^2(a)=1$ first on both sides then factorize.

$$\sin(x)[\sin(2x)-\sin(x)]=\sin(4x)[\sin(5x)-\sin(4x)]$$

Now use $\sin(A)−\sin(B)$ as a product on both sides and simplify (giving some roots $\sin(\frac{x}{2})=0$)

$$\sin(x)\cdot2\cos(\frac{3x}{2})\sin(\frac{x}{2})=\sin(4x)\cdot2\cos(\frac{9x}{2})\sin(\frac{x}{2})$$

then write \sin(C)\cos(D) as a difference of sines on both sides and simplify.

$$\frac{1}{2}\bigg[\sin(\frac{5x}{2})+\sin(\frac{-x}{2})\bigg]=\frac{1}{2}\bigg[\sin(\frac{17x}{2})+\sin(\frac{-x}{2})\bigg]$$

Write the difference of sines as a product and solve

$$\sin(\frac{5x}{2})-\sin(\frac{17x}{2})=0$$

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