Question

Find the number of solution of ${\sin }^{2}x-\sin x-1=0$ in $[-2\pi ,2\pi ]$

Answer 1

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

The above equation is quadratic in $\sin x$

$\therefore \sin x=\frac{-(-1)\pm \sqrt{{(-1)}^2-\mathrm{4.1.}(-1)}}{2.1}\Rightarrow \sin x=\frac{1+\sqrt{5}}{2}and\frac{1-\sqrt{5}}{2}Here,\sin \ne \frac{1+\sqrt{5}}{2}(\because \frac{1+\sqrt{5}}{2}>1)\therefore \sin x=\frac{1-\sqrt{5}}{2}$

Now, $\frac{1-\sqrt{5}}{2}$ is a negative value

If we observe the value of $\sin x$ takes all negative values less than $-1$ at two points

$\therefore \sin x=\frac{1-\sqrt{5}}{2}attwopintsin[0,2\pi ]also,\sin x=\frac{1-\sqrt{5}}{2}attwopointsin[-2\pi ,0]\therefore {sin}^{2}x-\sin x-1=0has4solutionsin[-2\pi ,2\pi ]$