Question

Solve the equation for the interval [0, 2π). 4 sin^2 x-1=0

Answer 1

$\bf 4sin^2(x)-1=0\qquad [0,2\pi ) \\\\\\ 4sin^2(x)=1\implies sin^2(x)=\cfrac{1}{4}\implies sin(x)=\pm\sqrt{\cfrac{1}{4}} \\\\\\ sin(x)=\pm\cfrac{\sqrt{1}}{\sqrt{4}}\implies sin(x)=\pm\cfrac{1}{2} \\\\\\ \measuredangle x= sin^{-1}\left( \pm\cfrac{1}{2} \right)\implies \measuredangle x= \begin{cases} \frac{\pi }{6}\\\\ \frac{5\pi }{6}\\\\ \frac{7\pi }{6}\\\\ \frac{11\pi }{6} \end{cases}$