One of the solutions of the equation 8 sin^3 | vedclass

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

Question

(B) Given equation: $8 \sin^3 	heta - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ Using the identity $4 \sin^3 	heta = 3 \sin 	heta - \sin 3 	heta$,

Vedclass · Accepted Answer

$\left(10^{\circ}, 20^{\circ}\right)$

Vedclass · Answer

(B) Given equation: $8 \sin^3 	heta - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ Using the identity $4 \sin^3 	heta = 3 \sin 	heta - \sin 3 	heta$,we substitute: $2(3 \sin 	heta - \sin 3 	heta) - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ $6 \sin 	heta - 2 \sin 3 	heta - 7 \sin 	heta + \sqrt{3} \cos 	heta = 0$ $\sqrt{3} \cos 	heta - \sin 	heta = 2 \sin 3 	heta$ Dividing by $2$: $\frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta = \sin 3 	heta$ $\sin(60^{\circ} - 	heta) = \sin 3 	heta$ $60^{\circ} - 	heta = 3 	heta \implies 4 	heta = 60^{\circ} \implies 	heta = 15^{\circ}$ Since $15^{\circ}$ lies in the interval $(10^{\circ}, 20^{\circ})$,the correct option is $B$.

One of the solutions of the equation $8 \sin^3 \theta - 7 \sin \theta + \sqrt{3} \cos \theta = 0$ lies in the interval

Similar Questions

If $0 \leq x \leq 2 \pi$,then the number of real values of $x$ which satisfy the equation $\sin x + \sin 2x + \sin 3x + \sin 4x = 0$ is

If $|k| = 5$ and $0^o \le \theta \le 360^o$,then the number of different solutions of $3\cos \theta + 4\sin \theta = k$ is

The equation $\sqrt{3} \sin x + \cos x = 4$ has

The number of values of $x$ in the interval $\left(\frac{\pi}{4}, \frac{7 \pi}{4}\right)$ for which $14 \operatorname{cosec}^{2} x - 2 \sin^{2} x = 21 - 4 \cos^{2} x$ holds,is

The number of solutions of $\sec x \cos 5x + 1 = 0$ in the interval $[0, 2\pi]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module