The number of solutions of the equation cos 2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equation: $\cos 2 	heta \cos \frac{	heta}{2} + \cos \frac{5 	heta}{2} = 2 \cos^3 \frac{5 	heta}{2}$ Using the identity $2 \cos^3 A - \co

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Given equation: $\cos 2 	heta \cos \frac{	heta}{2} + \cos \frac{5 	heta}{2} = 2 \cos^3 \frac{5 	heta}{2}$ Using the identity $2 \cos^3 A - \cos A = \cos 3A$,we rewrite the equation as: $\cos 2 	heta \cos \frac{	heta}{2} = 2 \cos^3 \frac{5 	heta}{2} - \cos \frac{5 	heta}{2} = \cos \left(3 	imes \frac{5 	heta}{2}ight) = \cos \frac{15 	heta}{2}$ Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,the $LHS$ becomes: $\frac{1}{2} \left( \cos \frac{5 	heta}{2} + \cos \frac{3 	heta}{2} ight) = \cos \frac{15 	heta}{2}$ This approach is complex; let us use the identity $\cos 3A = 4 \cos^3 A - 3 \cos A$. Actually,the equation simplifies to $\cos \frac{3 	heta}{2} = \cos \frac{15 	heta}{2}$. This implies $\frac{15 	heta}{2} = 2n \pi \pm \frac{3 	heta}{2}$. Case $1$: $\frac{15 	heta}{2} = 2n \pi + \frac{3 	heta}{2} \implies 6 	heta = 2n \pi \implies 	heta = \frac{n \pi}{3}$. For $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$	heta \in \{-\frac{\pi}{3}, 0, \frac{\pi}{3}\}$. Case $2$: $\frac{15 	heta}{2} = 2n \pi - \frac{3 	heta}{2} \implies 9 	heta = 2n \pi \implies 	heta = \frac{2n \pi}{9}$. For $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,$	heta \in \{-\frac{4 \pi}{9}, -\frac{2 \pi}{9}, 0, \frac{2 \pi}{9}, \frac{4 \pi}{9}\}$. Combining both sets and removing duplicates $(0)$: $	heta \in \{-\frac{4 \pi}{9}, -\frac{\pi}{3}, -\frac{2 \pi}{9}, 0, \frac{2 \pi}{9}, \frac{\pi}{3}, \frac{4 \pi}{9}\}$. Total number of solutions is $7$.

The number of solutions of the equation $\cos 2 \theta \cos \frac{\theta}{2} + \cos \frac{5 \theta}{2} = 2 \cos^3 \frac{5 \theta}{2}$ in the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is:

Similar Questions

If $\cos \theta = - \frac{1}{\sqrt{2}}$ and $\tan \theta = 1$,then the general value of $\theta$ is

If $x \neq (2n+1) \frac{\pi}{4}$,then the general solution of $\cos x + \cos 3x = \sin x + \sin 3x$ is

If $\tan \theta + \tan 2\theta + \sqrt{3} \tan \theta \tan 2\theta = \sqrt{3},$ then

If the sum of all the solutions of the equation $8 \cos x \cdot \left( \cos \left( \frac{\pi}{6} + x \right) \cdot \cos \left( \frac{\pi}{6} - x \right) - \frac{1}{2} \right) = 1$ in the interval $[0, \pi]$ is $k\pi$,then $k$ is equal to:

The equation $3\sin^2 x + 10\cos x - 6 = 0$ is satisfied,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module