The number of values of x for which sin 2x + | vedclass

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

Question

(A) We are given the equation $\sin 2x + \sin 4x = 2$.Since the maximum value of the sine function is $1$,the equation $\sin 2x + \sin 4x = 2$ can onl

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We are given the equation $\sin 2x + \sin 4x = 2$.Since the maximum value of the sine function is $1$,the equation $\sin 2x + \sin 4x = 2$ can only hold if $\sin 2x = 1$ and $\sin 4x = 1$ simultaneously.For $\sin 2x = 1$,we have $2x = 2n\pi + \frac{\pi}{2}$,which implies $x = n\pi + \frac{\pi}{4}$ for any integer $n$.For $\sin 4x = 1$,we have $4x = 2m\pi + \frac{\pi}{2}$,which implies $x = \frac{m\pi}{2} + \frac{\pi}{8}$ for any integer $m$.If we substitute $x = n\pi + \frac{\pi}{4}$ into $\sin 4x$,we get $\sin(4(n\pi + \frac{\pi}{4})) = \sin(4n\pi + \pi) = \sin(\pi) = 0$.Since $0 
eq 1$,there is no value of $x$ that satisfies both equations simultaneously.Therefore,the number of values of $x$ is $0$.

The number of values of $x$ for which $\sin 2x + \sin 4x = 2$ is

Similar Questions

If $\tan (\cot x) = \cot (\tan x),$ then $\sin 2x =$

If $\sin 2\theta = \cos \theta$ and $0 < \theta < \pi$,then the possible values of $\theta$ are:

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

The general value of $\theta$ satisfying the equation $2\sin^2 \theta - 3\sin \theta - 2 = 0$ is

The number of values of $x$ in the interval $[0, 5\pi]$ satisfying the equation $3 \sin^2 x - 7 \sin x + 2 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module