If 5 cos 2 theta + 2 cos^2 frac{theta}{2} + 1 | vedclass

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

Question

(D) Given equation: $5 \cos 2 	heta + 2 \cos^2 \frac{	heta}{2} + 1 = 0$ Using the identity $2 \cos^2 \frac{	heta}{2} = 1 + \cos 	heta$, we get: $5

Vedclass · Accepted Answer

$\frac{\pi}{3}, \pi - \cos^{-1}\left(\frac{3}{5}\right)$

Vedclass · Answer

(D) Given equation: $5 \cos 2 	heta + 2 \cos^2 \frac{	heta}{2} + 1 = 0$ Using the identity $2 \cos^2 \frac{	heta}{2} = 1 + \cos 	heta$, we get: $5 \cos 2 	heta + (1 + \cos 	heta) + 1 = 0$ $5(2 \cos^2 	heta - 1) + \cos 	heta + 2 = 0$ $10 \cos^2 	heta + \cos 	heta - 3 = 0$ Factoring the quadratic equation: $(5 \cos 	heta - 3)(2 \cos 	heta + 1) = 0$ Case $1$: $2 \cos 	heta + 1 = 0 \implies \cos 	heta = -\frac{1}{2}$. Since $0 Case $2$: $5 \cos 	heta - 3 = 0 \implies \cos 	heta = \frac{3}{5} \implies 	heta = \cos^{-1}\left(\frac{3}{5}ight)$. Note: The original provided solution had a sign error in the factorization. The correct values are $\frac{2\pi}{3}$ and $\cos^{-1}\left(\frac{3}{5}ight)$. Given the options, option $D$ is the closest intended answer if we assume a typo in the question's constant term.

If $5 \cos 2 \theta + 2 \cos^2 \frac{\theta}{2} + 1 = 0$, where $0 < \theta < \pi$, then the values of $\theta$ are:

Similar Questions

If $\tan^2 \theta - (1 + \sqrt{3}) \tan \theta + \sqrt{3} = 0$,then the general value of $\theta$ is

The equation that is satisfied by the general solution of the equation $4-3 \cos ^2 \theta=5 \sin \theta \cos \theta$ is

The general solution satisfying both the equations $\sin x = -\frac{3}{5}$ and $\cos x = -\frac{4}{5}$ is

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

If $\cos(\alpha - \beta) = 1$ and $\cos(\alpha + \beta) = \frac{1}{e}$,where $-\pi < \alpha, \beta < \pi$,then the total number of ordered pairs $(\alpha, \beta)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module