Jika sin2(x−π6)=12\sin^2\left(x-\frac{\pi}{6}\right)=\frac{1}{2}sin2(x−6π)=21 untuk 0≤x≤2π0\le x\le2\pi0≤x≤2π maka nilai xxx yang memenuhi adalah ....
{5π12,11π12,17π12,23π12}\left\{\frac{5\pi}{12},\frac{11\pi}{12},\frac{17\pi}{12},\frac{23\pi}{12}\right\}{125π,1211π,1217π,1223π}
{5π13,9π13,17π13,21π13}\left\{\frac{5\pi}{13},\frac{9\pi}{13},\frac{17\pi}{13},\frac{21\pi}{13}\right\}{135π,139π,1317π,1321π}
{3π7,4π7,9π7,23π7}\left\{\frac{3\pi}{7},\frac{4\pi}{7},\frac{9\pi}{7},\frac{23\pi}{7}\right\}{73π,74π,79π,723π}
{5π12,11π12,17π12}\left\{\frac{5\pi}{12},\frac{11\pi}{12},\frac{17\pi}{12}\right\}{125π,1211π,1217π}
{3π7,9π7,23π7}\left\{\frac{3\pi}{7},\frac{9\pi}{7},\frac{23\pi}{7}\right\}{73π,79π,723π}