The number of roots of the equation cos^2 x + | vedclass

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

Question

(B) Given equation: $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$Substitute $\cos^2 x = 1 - \sin^2 x$:$1 - \sin^2 x + \frac{

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given equation: $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$Substitute $\cos^2 x = 1 - \sin^2 x$:$1 - \sin^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$$-\sin^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} = 0$Multiply by $-4$:$4 \sin^2 x - 2(\sqrt{3} + 1) \sin x + \sqrt{3} = 0$$4 \sin^2 x - 2\sqrt{3} \sin x - 2 \sin x + \sqrt{3} = 0$$2 \sin x (2 \sin x - \sqrt{3}) - 1(2 \sin x - \sqrt{3}) = 0$$(2 \sin x - 1)(2 \sin x - \sqrt{3}) = 0$So,$\sin x = \frac{1}{2}$ or $\sin x = \frac{\sqrt{3}}{2}$.For $\sin x = \frac{1}{2}$ in $[-\pi, \pi]$,$x = \frac{\pi}{6}, \frac{5\pi}{6}$.For $\sin x = \frac{\sqrt{3}}{2}$ in $[-\pi, \pi]$,$x = \frac{\pi}{3}, \frac{2\pi}{3}$.Total number of roots is $4$.

The number of roots of the equation $\cos^2 x + \frac{\sqrt{3} + 1}{2} \sin x - \frac{\sqrt{3}}{4} - 1 = 0$ that lie in the interval $[-\pi, \pi]$ is:

Similar Questions

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

The general solution of the trigonometric equation $(\sqrt{3}-1) \sin \theta + (\sqrt{3}+1) \cos \theta = 2$ is

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

If $\cot \theta = \sin 2\theta$ (where $\theta \neq n\pi$,$n$ is an integer),then $\theta$ is equal to:

If $\tan m\theta = \tan n\theta$,then the general values of $\theta$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module