Untuk −π≤x≤π-\pi\le x\le\pi−π≤x≤π nilai xxx yang memenuhi persamaan 4cos2x−4sin(π2+x)−3=04\cos^2x-4\sin\left(\frac{\pi}{2}+x\right)-3=04cos2x−4sin(2π+x)−3=0 adalah ....
{−2π3,2π3}\left\{-\frac{2\pi}{3},\frac{2\pi}{3}\right\}{−32π,32π}
{−2π5,2π5}\left\{-\frac{2\pi}{5},\frac{2\pi}{5}\right\}{−52π,52π}
{1π3,1π2}\left\{\frac{1\pi}{3},\frac{1\pi}{2}\right\}{31π,21π}
{−2π5,2π7}\left\{-\frac{2\pi}{5},\frac{2\pi}{7}\right\}{−52π,72π}
{2π5,3π5}\left\{\frac{2\pi}{5},\frac{3\pi}{5}\right\}{52π,53π}