Question

Find the number of solutions of ${\sin }^2-\sin x-1=0$ in the interval $(\frac{-\pi }{2},0)$.

Answer 1

Given,

${\sin }^2\left(x\right)-\sin \left(x\right)-1=0$

Let, $u=\sin x$

$\Rightarrow u^2-u-1=0$

$u=\frac{1+\sqrt{5}}{2},u=\frac{1-\sqrt{5}}{2}$

$\Rightarrow \sin \left(x\right)=\frac{1+\sqrt{5}}{2},\sin \left(x\right)=\frac{1-\sqrt{5}}{2}$

$\Rightarrow \sin \left(x\right)=\frac{1+\sqrt{5}}{2}>1$ but $-1<\sin x<1$

$\therefore Now,x={\sin }^{-1}\left(\frac{1-\sqrt{5}}{2}\right)$

$=-{38.17}^{\circ }$

Hence the equation has only $1$ root.