Solve for $x$: $2\sin^2(x) - \sin^2(2x) = \cos^2(2x)$

Gonna repeat it in the body for readability, because I don't know if markers work in the title

$2\sin^2(x) - \sin^2(2x) = \cos^2(2x)$

I also put it into Wolfram Alpha and it returned relatively incomprehensible answer involving polynomial transformations, if possible a simpler answer would be highly appreciated.

Solution 1:

Your equation is simply

$$2\sin^2(x) = \underbrace{\cos^2(2x) + \sin^2(2x)}_{=1}$$

Whence

$$\sin^2(x) = \frac{1}{2}$$

Can you proceed from here?

Solution

$$\sin(x) = \pm \frac{\sqrt{2}}{2}$$

This is a well known value and it's $\pi/4$ radians and $3\pi/4$ radians. Since we have $\sin^2$, to be then explicit in the solution we have also $-\pi/4$ and $5\pi/4$ radians:

$$x = \pm \frac{\pi}{4} + 2k\pi ~~~~~~~~~~~ k\in\mathbb{Z}$$ $$x = \frac{3\pi}{4} + 2k\pi ~~~~~~~~~~~ k\in\mathbb{Z}$$ $$x = \frac{5\pi}{4} + 2k\pi ~~~~~~~~~~~ k\in\mathbb{Z}$$