Question

Solve:${\cos }^{2}x-\sin x-\frac{1}{4}=0$

Answer 1

${\cos }^{2}x-\sin x-\frac{1}{4}=0$

Replacing ${\cos }^{2}x$ by $1-{\sin }^{2}x$ we get a quadratic in $sine$ as

$4{\sin }^{2}x+4\sin x-3=0$

$\Rightarrow 4{\sin }^{2}x+6\sin x-2\sin x-3=0$

$\Rightarrow 2\sin x(2\sin x+3)-1(2\sin x+3)=0$

$\Rightarrow (2\sin x+3)(2\sin x-1)=0$

$\Rightarrow \sin x\ne \frac{-3}{2}$ ince $-1\le \sin x\le 1$

If $\sin x=\frac{1}{2}\Rightarrow$Principal solution is $x=\frac{\pi }{6}$

General solution is $x=n\pi +{(-1)}^n\frac{\pi }{6},n\in I$