The solution set of (5+4 cos theta)(2 cos the | vedclass

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

Question

(C) Given the equation: $(5+4 \cos 	heta)(2 \cos 	heta+1)=0$Since $5+4 \cos 	heta$ can never be $0$ (because $-1 \le \cos 	heta \le 1$,so $1 \le 5

Vedclass · Accepted Answer

$\left\{\frac{2 \pi}{3}, \frac{4 \pi}{3}\right\}$

Vedclass · Answer

(C) Given the equation: $(5+4 \cos 	heta)(2 \cos 	heta+1)=0$Since $5+4 \cos 	heta$ can never be $0$ (because $-1 \le \cos 	heta \le 1$,so $1 \le 5+4 \cos 	heta \le 9$),we must have:$2 \cos 	heta + 1 = 0$$\cos 	heta = -\frac{1}{2}$In the interval $[0, 2\pi]$,the values of $	heta$ for which $\cos 	heta = -\frac{1}{2}$ are:$	heta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $	heta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$Thus,the solution set is $\left\{\frac{2\pi}{3}, \frac{4\pi}{3}ight\}$.

The solution set of $(5+4 \cos \theta)(2 \cos \theta+1)=0$ in the interval $[0, 2\pi]$ is:

Similar Questions

The solution of the equation $\sec \theta - \text{cosec} \theta = \frac{4}{3}$ is

The equation $\sin^2 \theta - \frac{4}{\sin^3 \theta - 1} = 1 - \frac{4}{\sin^3 \theta - 1}$ has:

Solve $\tan(x) + \sec(x) = \sqrt{3}$ for $x \in [0, 2\pi]$.

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

The general solution of the equation $\tan x + \tan 2x - \tan 3x = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module