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: $$ \cos^2(x)- \cos^2(4x)= \sin(5x)\sin(3x) $$
Best Answer
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
$$\sin(\frac{5x}{2})-\sin(\frac{17x}{2})=0$$