Question

Please help me solve this,

Answer 1

$\cos^2\theta=1-\sin^2\theta$

$2\cos^2\theta=\sin\theta+1$
$2-2\sin^2\theta=\sin\theta+1$
$2\sin^2\theta+\sin\theta-1=0$
$(2\sin\theta-1)(\sin\theta+1)=0$

$2\sin\theta-1=0\implies \sin\theta=\dfrac12\implies\theta=\dfrac\pi6,\dfrac{5\pi}6$
$\sin\theta+1=0\implies\sin\theta=-1\implies\theta=\dfrac{3\pi}2$