The set of solutions of the equation (sqrt{3} | vedclass

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

Question

(A) Given equation: $(\sqrt{3}-1) \sin 	heta+(\sqrt{3}+1) \cos 	heta=2$Divide by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3+1-2\sqrt{3} + 3+1

Vedclass · Accepted Answer

$\{2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}: n \in Z\}$

Vedclass · Answer

(A) Given equation: $(\sqrt{3}-1) \sin 	heta+(\sqrt{3}+1) \cos 	heta=2$Divide by $\sqrt{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2} = \sqrt{3+1-2\sqrt{3} + 3+1+2\sqrt{3}} = \sqrt{8} = 2\sqrt{2}$.$\frac{\sqrt{3}-1}{2\sqrt{2}} \sin 	heta + \frac{\sqrt{3}+1}{2\sqrt{2}} \cos 	heta = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}$.We know $\cos 15^{\circ} = \cos \frac{\pi}{12} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ and $\sin 15^{\circ} = \sin \frac{\pi}{12} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.So,$\sin \frac{\pi}{12} \sin 	heta + \cos \frac{\pi}{12} \cos 	heta = \cos \frac{\pi}{4}$.$\cos(	heta - \frac{\pi}{12}) = \cos \frac{\pi}{4}$.General solution: $	heta - \frac{\pi}{12} = 2n\pi \pm \frac{\pi}{4}$.$	heta = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$,where $n \in Z$.

The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

Similar Questions

The solution set of the equation $\tan x + \sec x = 2 \cos x$ in the interval $[0, 2 \pi]$ is

The number of solutions of $\cos 2 \theta = \sin \theta$ in the interval $(0, 2 \pi)$ is:

If $\frac{1}{6} \sin \theta, \cos \theta, \tan \theta$ are in $G.P.$,then the solution set of $\theta$ is

The sum of the solutions of $\cos x \sqrt{16 \sin ^2 x} = 1$ in $(0, 2 \pi)$ is

The general solution of $|\sin x|=\cos x$ is (when $n \in I$) given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module