Find the general solution of sin x = -frac{sq | vedclass

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

Question

(A) Given $\sin x = -\frac{\sqrt{3}}{2}$.Since $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $\sin x = -\sin \frac{\pi}{3}$.Using the identity $\s

Vedclass · Accepted Answer

$x = n\pi + (-1)^n \frac{4\pi}{3}, n \in \mathbb{Z}$

Vedclass · Answer

(A) Given $\sin x = -\frac{\sqrt{3}}{2}$.Since $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$,we have $\sin x = -\sin \frac{\pi}{3}$.Using the identity $\sin(\pi + 	heta) = -\sin 	heta$,we get $\sin x = \sin(\pi + \frac{\pi}{3}) = \sin \frac{4\pi}{3}$.The general solution for $\sin x = \sin \alpha$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$.Substituting $\alpha = \frac{4\pi}{3}$,we get $x = n\pi + (-1)^n \frac{4\pi}{3}$,where $n \in \mathbb{Z}$.

Find the general solution of $\sin x = -\frac{\sqrt{3}}{2}$.

Similar Questions

The number of solutions of the equation $(4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\frac{4}{1+\sqrt{3}}$ for $x \in [-2\pi, \frac{5\pi}{2}]$ is

The solution set of the trigonometric equation $\tan \theta + 5 \cot \theta = \sec \theta$ is

If $n$ is the number of solutions of the equation $2 \cos x(4 \sin(\frac{\pi}{4}+x) \sin(\frac{\pi}{4}-x)-1)=1$ for $x \in [0, \pi]$,and $S$ is the sum of all these solutions,then the ordered pair $(n, S)$ is:

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:

In $(0, 2\pi)$,the number of solutions of $\tan \theta + \sec \theta = 2 \cos \theta$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module