The sum of solutions in x in (0, 2pi) of the | vedclass

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

Question

(A) Using the identity $\cos(60^{\circ} - x)\cos(60^{\circ} + x) = \cos^2(60^{\circ}) - \sin^2(x) = \frac{1}{4} - \sin^2(x)$.Substituting this into th

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Using the identity $\cos(60^{\circ} - x)\cos(60^{\circ} + x) = \cos^2(60^{\circ}) - \sin^2(x) = \frac{1}{4} - \sin^2(x)$.Substituting this into the equation: $4\cos(x) \left(\frac{1}{4} - \sin^2(x)\right) = 1$.$4\cos(x) \left(\frac{1 - 4\sin^2(x)}{4}\right) = 1$.$\cos(x)(1 - 4\sin^2(x)) = 1$.$\cos(x)(1 - 4(1 - \cos^2(x))) = 1$.$\cos(x)(4\cos^2(x) - 3) = 1$.$4\cos^3(x) - 3\cos(x) = 1$.Using the identity $\cos(3x) = 4\cos^3(x) - 3\cos(x)$, we get $\cos(3x) = 1$.For $x \in (0, 2\pi)$, $3x \in (0, 6\pi)$.$3x = 2\pi, 4\pi$.$x = \frac{2\pi}{3}, \frac{4\pi}{3}$.The sum of the solutions is $\frac{2\pi}{3} + \frac{4\pi}{3} = \frac{6\pi}{3} = 2\pi$.

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 $pi$)

Similar Questions

The general solution of $\sin^{2} x \cdot \sec x = \tan x - \sin x + 1$ is

The general solution of $2 \sqrt{3} \cos^2 \theta = \sin \theta$ is

The number of elements in the set $\{x \in [0, 180^{\circ}] : \tan(x+100^{\circ}) = \tan(x+50^{\circ}) \tan x \tan(x-50^{\circ})\}$ is . . . . . . .

If the general solution of $\sin x + 3 \sin 3x + \sin 5x = 0$ is $x = y$,then the set of all values of $\cos y$ is

The general solution of $\sin x - \cos x = \sqrt{2}$,for any integer $n$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module