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

Using $\displaystyle\cos2A=1-2\sin^2A,\cos4x=0\implies 4x=(2n+1)\frac\pi2$ where $n$ is an integer
