If (2cos x - 1)(3 + 2cos x) = 0 for 0 le x le | vedclass

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

Question

(B) Given the equation $(2\cos x - 1)(3 + 2\cos x) = 0$. This implies either $2\cos x - 1 = 0$ or $3 + 2\cos x = 0$. Case $1$: $2\cos x - 1 = 0 \Right

Vedclass · Accepted Answer

$\frac{\pi}{3}, \frac{5\pi}{3}$

Vedclass · Answer

(B) Given the equation $(2\cos x - 1)(3 + 2\cos x) = 0$. This implies either $2\cos x - 1 = 0$ or $3 + 2\cos x = 0$. Case $1$: $2\cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}$. For $0 \le x \le 2\pi$,the solutions are $x = \frac{\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. Case $2$: $3 + 2\cos x = 0 \Rightarrow \cos x = -\frac{3}{2}$. Since the range of $\cos x$ is $[-1, 1]$,the equation $\cos x = -\frac{3}{2}$ has no real solution. Thus,the values of $x$ in the interval $[0, 2\pi]$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

If $(2\cos x - 1)(3 + 2\cos x) = 0$ for $0 \le x \le 2\pi$,then $x = $

Similar Questions

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

The sum of the solutions in $(0, 2\pi)$ for the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right) = \frac{1}{4}$ is

The number of solutions of the equation $\tan x + \sec x = 2 \cos x$ lying in the interval $[0, 2 \pi]$ is

If $0 < \theta < \frac{\pi}{2}$,then the solution of the equation $\sin \theta - 3 \sin 2 \theta + \sin 3 \theta = \cos \theta - 3 \cos 2 \theta + \cos 3 \theta$ is

If $\sin \theta + \cos \theta = 0$ and $0 < \theta < \pi$, then $\theta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module