If \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] is equal to
Options
1
-1
- \[- \sqrt{5}\]
- \[\sqrt{5}\]
Solution
1
\[We have, \]
\[\cos2x + 2\text{ cos } x = 1\]
\[ \Rightarrow 2 \cos^2 x - 1 + 2\text{ cos } x = 1\]
\[ \Rightarrow \cos^2 x + \text{ cos } x - 1 = 0\]
\[ \Rightarrow \text{ cos } x = \frac{- 1 \pm \sqrt{1^2 + 4}}{2}\]
\[ \Rightarrow \text{ cos } x = \frac{- 1 \pm \sqrt{5}}{2}\]
\[ \Rightarrow \text{ cos } x = \frac{- 1 + \sqrt{5}}{2}\]
\[\text{ Now, } \]
\[\left( 2 - \cos^2 x \right) \sin^2 x = \left[ 2 - \left( \frac{- 1 + \sqrt{5}}{2} \right)^2 \right] \left( 1 - \cos^2 x \right)\]
\[ = \left[ 2 - \frac{1}{4}\left( 1 - 2\sqrt{5} + 5 \right) \right] \left( 1 - \frac{1}{4}\left( 1 - 2\sqrt{5} + 5 \right) \right)\]
\[ = \frac{1}{4}\left( 1 + \sqrt{5} \right)\left( \sqrt{5} - 1 \right) = \frac{4}{4} = 1\]
