The values of $4 {\sin }^2 \theta - \cos 2 \theta$ lies in the interval

Solution:

Let $x=4{\sin }^2\theta -\cos 2\theta$. Then,

   $\begin{array}{l} x=2(1-\cos 2\theta )-\cos 2\theta \\ \Rightarrow x=2-3\cos 2\theta \end{array}$

Now,

     $-3\le -3\cos 2\theta \le 3$                                $[\because -1\le \cos 2\theta \le 1]$

$\begin{array}{l}\Rightarrow 2-3\le 2-3\cos 2\theta \le 2+3\\ \Rightarrow -1\le x\le 5\Rightarrow x\in [-1,5].\end{array}$