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{√(1)}{√(4)}\implies sin(x)=\pm\cfrac{1}{2} \\\\\\ \measuredangle x= sin^(-1)\left( \pm\cfrac{1}{2} \right)\implies \measuredangle x= \begin{cases} (\pi )/(6)\\\\ (5\pi )/(6)\\\\ (7\pi )/(6)\\\\ (11\pi )/(6) \end{cases}`