If cos 2x + 2 cos x = 1, Find the Value of (2 − cos²x) sin²x
Question
If
\[ \cos 2x + 2\cos x = 1 \]
then find the value of
\[ (2-\cos^2x)\sin^2x \]
(a) \(1\)
(b) \(-1\)
(c) \(-\sqrt5\)
(d) \(\sqrt5\)
Solution
Using the identity
\[ \cos 2x = 2\cos^2x-1 \]
Substitute in the given equation:
\[ 2\cos^2x-1+2\cos x=1 \]
\[ 2\cos^2x+2\cos x-2=0 \]
\[ \cos^2x+\cos x-1=0 \]
Let \[ t=\cos x \] Then
\[ t^2+t-1=0 \]
Hence,
\[ t^2=1-t \]
and
\[ \sin^2x=1-\cos^2x =1-(1-t) =t \]
Now,
\[ (2-\cos^2x)\sin^2x = (2-t^2)t \]
Using \(t^2=1-t\),
\[ =(2-(1-t))t \]
\[ =(1+t)t \]
\[ =t+t^2 \]
Again using \[ t^2+t=1 \]
\[ (2-\cos^2x)\sin^2x=1 \]
Final Answer
\[ \boxed{1} \]
Therefore, the correct option is (a) 1.