Pre-cal trigonometric equation problem am i correct to factor out a 2 first?
$$2\sin^2 2x =1$$
$$2(\sin^2 x-1) =1$$
$$2\cos^2 x =1$$
$$\cos^2 x ={1\over2}$$
$$\cos x =\pm{\sqrt{2}\over 2 }$$
i'm only looking for solutions from $$0≤ x ≤ 2\pi $$
$$x = {\frac{\pi}{4}},{\frac{7\pi}{4}},{\frac{3\pi}{4}},{\frac{5\pi}{4}}$$
thanks for any corrections

Hint: let $y = \sin 2x$ and solve for $y$ first using algebra. Then figure out what value(s) of $x$ would make the equation true.
