Find general solution of the given trigonometric equation:
$\sin^24x + \cos^2x = 2 \sin4x \cos^2x$
I tried converting the whole equation in the form of $2x$ and got a pretty complicated equation involving $\sin2x, \cos2x, \sin^22x$ and $\cos^22x$. I have got no idea how to further solve that equation.
Best Answer
Rearrange the equation as follows,
$$\sin^24x + \cos^2x - 2 \sin4x \cos^2x$$ $$=(\cos^2x - \sin 4x)^2 -\cos^4 x+\cos^2x$$ $$=(\cos^2x - \sin 4x)^2 +\cos^2x \sin^2x$$ $$=(\cos^2x - 2\sin 2x\cos2x)^2 +\frac14\sin^2 2x=0$$
where both terms vanish, leading to the following system of equations, $$\sin2x =0$$ $$\cos^2x=2\cos2x\sin2x=0$$
Thus, the valid solutions are $x=\frac\pi2+n\pi$.