Question:
If $\cos 2 x+2 \cos x=1$ then, $\left(2-\cos ^{2} x\right) \sin ^{2} x$ is equal to
(a) 1
(b) $-1$
(c) $-\sqrt{5}$
(d) $\sqrt{5}$
Solution:
(a) 1
We have,
$\cos 2 x+2 \cos x=1$
$\Rightarrow 2 \cos ^{2} x-1+2 \cos x=1$
$\Rightarrow \cos ^{2} x+\cos x-1=0$
$\Rightarrow \cos x=\frac{-1 \pm \sqrt{1^{2}+4}}{2}$
$\Rightarrow \cos x=\frac{-1 \pm \sqrt{5}}{2}$
$\Rightarrow \cos x=\frac{-1+\sqrt{5}}{2}$
Now,
$\left(2-\cos ^{2} x\right) \sin ^{2} x=\left[2-\left(\frac{-1+\sqrt{5}}{2}\right)^{2}\right]\left(1-\cos ^{2} x\right)$
$=\left[2-\frac{1}{4}(1-2 \sqrt{5}+5)\right]\left(1-\frac{1}{4}(1-2 \sqrt{5}+5)\right)$
$=\frac{1}{4}(1+\sqrt{5})(\sqrt{5}-1)=\frac{4}{4}=1$