Find the number of solutions of the equation $\sin^22x-\cos^28x=\frac{1}{2}\cos10x$ lying in the interval $(0,\frac{\pi}{2})$
I found the period of $\sin^22x-\cos^28x$ as $\pi$ and the period of $\frac{1}{2}\cos10x$ is $\frac{\pi}{5}$
I do not know how to solve it further.
$$\sin^2 2x-\cos^2 8x=\frac{1}{2} \cos10x$$
Using the formulae, $$\sin^2\alpha=\frac{1-\cos 2 \alpha}{2}; \cos^2\alpha=\frac{1+\cos 2 \alpha}{2}$$
$$\frac{1-\cos 4x}{2}-\frac{1+\cos 16x}{2}=\frac{1}{2} \cos10x$$ $$-\cos 4x-\cos 16 x=\cos 10 x$$ $$2 \cos 10x \cos 6x + \cos 10x=0$$ $\cos 10x =0$ or $\cos 6x= -\frac 12$
HINT:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$\cos^28x-\sin^22x=\cos10x\cos6x$$
So, $$\cos^28x-\sin^22x=-\dfrac{\cos10x}2$$
$$\implies\cos10x=-2(\cos^28x-\sin^22x)=-2\cos10x\cos6x$$ $$\iff\cos10x(2\cos6x+1)=0$$
Hope you can take it from here
