Question

If $\cos 2x+2\cos x=1$ then, $\left(2-{\cos }^{2}x\right){\sin }^{2}x$ is equal to
(a) 1

(b) $-1$

(c) $-\sqrt{5}$

(d) $\sqrt{5}$

Answer 1

(a) 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \cos 2x+2\cos x=1 \\ \Rightarrow 2{\cos }^{2}x-1+2\cos x=1 \\ \Rightarrow {\cos }^{2}x+\cos x\mathit{-}1=0 \\ \Rightarrow \cos x=\frac{-1\pm \sqrt{1^2+4}}{2} \\ \Rightarrow \cos x=\frac{-1\pm \sqrt{5}}{2} \\ \Rightarrow \cos x=\frac{-1+\sqrt{5}}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{Now}, \\ \left(2-{\cos }^2\mathrm{x}\right){\sin }^{2}x=\left[2-{\left(\frac{-1+\sqrt{5}}{2}\right)}^2\right]\left(1-{\cos }^{2}x\right) \\ =\left[2-\frac{1}{4}\left(1-2\sqrt{5}+5\right)\right]\left(1-\frac{1}{4}\left(1-2\sqrt{5}+5\right)\right) \\ =\frac{1}{4}\left(1+\sqrt{5}\right)\left(\sqrt{5}-1\right)=\frac{4}{4}=1 \end{gathered}$$