Question

Solve the equation on the
interval [0, 27r).
4(sin x)2 - 2 = 0

Answer 1

`4[sin(x)]^2 - 2 = 0\implies 4[sin(x)]^2=2\implies [sin(x)]^2=\cfrac{2}{4}\implies [sin(x)]^2=\cfrac{1}{2} \\\\\\ sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies sin^(-1)[sin(x)]=sin^(-1)\left( \pm\sqrt{\cfrac{1}{2}} \right)\implies x=sin^(-1)\left( \pm\sqrt{\cfrac{1}{2}} \right)`

`x=sin^(-1)\left( \pm\cfrac{√(1)}{√(2)} \right)\implies x=sin^(-1)\left( \pm\cfrac{1}{√(2)} \right)\implies x=sin^(-1)\left( \pm\cfrac{√(2)}{2} \right) \\\\[-0.35em] ~\dotfill\\\\ ~\hfill x=\cfrac{\pi }{4}~~,~~\cfrac{3\pi }{4}~~,~~\cfrac{5\pi }{4}~~,~~\cfrac{7\pi }{4}~\hfill`

Check the picture below.