Let S={x in(-pi, pi) mid x neq 0, pm frac{pi} | vedclass

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

Question

(C) Given equation: $\sqrt{3} \sec x+\operatorname{cosec} x+2(	an x-\cot x)=0$Dividing by $2$: $\frac{\sqrt{3}}{2} \sec x+\frac{1}{2} \operatorname{c

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given equation: $\sqrt{3} \sec x+\operatorname{cosec} x+2(	an x-\cot x)=0$Dividing by $2$: $\frac{\sqrt{3}}{2} \sec x+\frac{1}{2} \operatorname{cosec} x=\cot x-	an x$Converting to $\sin x$ and $\cos x$: $\frac{\sqrt{3}}{2 \cos x}+\frac{1}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$Multiplying by $\sin x \cos x$: $\frac{\sqrt{3}}{2} \sin x+\frac{1}{2} \cos x=\cos^2 x-\sin^2 x$Using $\cos(A-B)$ and $\cos 2x$ formulas: $\cos(x-\frac{\pi}{6})=\cos 2x$General solution: $2x = 2n\pi \pm (x-\frac{\pi}{6})$Case $1$: $2x = 2n\pi + x - \frac{\pi}{6} \implies x = 2n\pi - \frac{\pi}{6}$For $n=0, x=-\frac{\pi}{6} \in S$. For $n=1, x=\frac{11\pi}{6} 
otin S$.Case $2$: $2x = 2n\pi - (x - \frac{\pi}{6}) \implies 3x = 2n\pi + \frac{\pi}{6} \implies x = \frac{2n\pi}{3} + \frac{\pi}{18}$For $n=0, x=\frac{\pi}{18} \in S$. For $n=1, x=\frac{13\pi}{18} \in S$. For $n=-1, x=-\frac{11\pi}{18} \in S$.Sum of solutions: $-\frac{\pi}{6} + \frac{\pi}{18} + \frac{13\pi}{18} - \frac{11\pi}{18} = \frac{-3\pi + \pi + 13\pi - 11\pi}{18} = 0$.

Let $S=\{x \in(-\pi, \pi) \mid x \neq 0, \pm \frac{\pi}{2}\}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

Similar Questions

In a triangle $ABC$,with usual notations,$\frac{\cos B+\cos C}{b+c}+\frac{\cos A}{a}$ has the value

If $r_1, r_2$ and $r_3$ of a $\triangle ABC$ are in Harmonic progression,then $a, b$ and $c$ will be in

The number of solutions to $\sin(\pi \sin^2 \theta) + \sin(\pi \cos^2 \theta) = 2 \cos(\frac{\pi}{2} \cos \theta)$ satisfying $0 \leq \theta \leq 2\pi$ is

Which of the following is true?

If $A, B, C, D$ are angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module