Question

The number of solution of ${\sin }^4\theta -2{\sin }^2\theta -1=0$ for $\theta \in [0,2\pi ]$ is

Answer 1

Given: ${\sin }^4\theta -2{\sin }^2\theta -1=0$
$\Rightarrow ({\sin }^2\theta -1{)}^2-2=0\Rightarrow {\sin }^2\theta =1\pm \sqrt{2}$
As ${\sin }^2\theta \in [0,1]$, so no solution of the given equation.