Number of solution(s) of the equation (cos2θ).(cos θ/2) + cos 5θ/2 = 2cos^3(5θ/2) in [-π/2, π/2] is equal to

Question

Number of solution(s) of the equation

\((\cos 2 \theta) \cdot\left(\cos \frac{\theta}{2}\right)+\cos \frac{5 \theta}{2}=2 \cos ^{3}\left(\frac{5 \theta}{2}\right)\) in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is equal to

(1) 6

(2) 7

(3) 4

(4) 2  

Answer 1

Correct option is: (2) 7

\(2(\cos 2 \theta) \cdot\left(\cos \frac{\theta}{2}\right)+2 \cos \frac{5 \theta}{2}=4 \cos ^{3}\left(\frac{5 \theta}{2}\right)\)

\(\Rightarrow \cos \left(\frac{5 \theta}{2}\right)+\cos \frac{3 \theta}{2}+2 \cos \left(\frac{5 \theta}{2}\right) =\left(\cos \frac{15 \theta}{2}+3 \cos \frac{5 \theta}{2}\right) \)

\(\Rightarrow \cos \left(\frac{3 \theta}{2}\right)+\cos \left(\frac{15 \theta}{2}\right)\)

\( \Rightarrow \cos \left(\frac{3 \theta}{2}\right)-\cos \frac{15 \theta}{2}=0 \)

\( \Rightarrow 2 \sin \left(\frac{9 \theta}{2}\right) \sin \left(\frac{6 \theta}{2}\right)=0,3 \theta=2 A \pi \)

\( \therefore \frac{9 \theta}{2}=\eta \pi \rightarrow \theta=\frac{2 \eta \pi}{9} \)

\( \Rightarrow \theta=\frac{2 \eta \pi}{3} \)

\(\therefore \theta=-\frac{4 \pi}{9},-\frac{3 \pi}{9},-\frac{2 \pi}{9}, 0, \frac{2 \pi}{9}, \frac{3 \pi}{9}, \frac{4 \pi}{9}\)