If sin x cos x = frac{1}{4}, then the general | vedclass

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

Question

(A) Given the equation $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$ to use the identity $\sin(2x) = 2 \sin x \cos x$:
$2 \sin x \cos x =

Vedclass · Accepted Answer

$\frac{n\pi}{2} + (-1)^n \frac{\pi}{12}$

Vedclass · Answer

(A) Given the equation $\sin x \cos x = \frac{1}{4}$.
Multiply both sides by $2$ to use the identity $\sin(2x) = 2 \sin x \cos x$:
$2 \sin x \cos x = 2 	imes \frac{1}{4}$
$\sin(2x) = \frac{1}{2}$
We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$.
The general solution for $\sin 	heta = \sin \alpha$ is $	heta = n\pi + (-1)^n \alpha$.
Here, $	heta = 2x$ and $\alpha = \frac{\pi}{6}$.
So, $2x = n\pi + (-1)^n \frac{\pi}{6}$.
Dividing by $2$, we get:
$x = \frac{n\pi}{2} + (-1)^n \frac{\pi}{12}$.

If $\sin x \cos x = \frac{1}{4}$, then the general solution is:

Similar Questions

If the equation in variable $\theta$,$3 \tan(\theta - \alpha) = \tan(\theta + \alpha)$,(where $\alpha$ is a constant) has no real solution,then $\alpha$ can be (wherever $\tan(\theta - \alpha)$ and $\tan(\theta + \alpha)$ are both defined).

If $\sin 2\theta = \cos 3\theta$ and $\theta$ is an acute angle,then $\sin \theta$ is equal to

If $\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$,where $0 < \theta < 180^\circ$,then $\theta = $

If $\tan 2\theta \tan \theta = 1$,then the general value of $\theta$ is

The general value of $\theta$ in the equation $2\sqrt{3} \cos \theta = \tan \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module