The number of distinct solutions of sec theta | vedclass

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

Question

(A) Given equation: $\sec 	heta + 	an 	heta = \sqrt{3}$$\Rightarrow \frac{1}{\cos 	heta} + \frac{\sin 	heta}{\cos 	heta} = \sqrt{3}$$\Rightarrow

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given equation: $\sec 	heta + 	an 	heta = \sqrt{3}$$\Rightarrow \frac{1}{\cos 	heta} + \frac{\sin 	heta}{\cos 	heta} = \sqrt{3}$$\Rightarrow 1 + \sin 	heta = \sqrt{3} \cos 	heta$ (where $\cos 	heta 
eq 0$)$\Rightarrow \sqrt{3} \cos 	heta - \sin 	heta = 1$Dividing by $2$ on both sides:$\Rightarrow \frac{\sqrt{3}}{2} \cos 	heta - \frac{1}{2} \sin 	heta = \frac{1}{2}$$\Rightarrow \cos 	heta \cos \frac{\pi}{6} - \sin 	heta \sin \frac{\pi}{6} = \cos \frac{\pi}{3}$$\Rightarrow \cos \left(	heta + \frac{\pi}{6}ight) = \cos \frac{\pi}{3}$General solution: $	heta + \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{3}, n \in Z$Case $1$: $	heta = 2n\pi + \frac{\pi}{3} - \frac{\pi}{6} = 2n\pi + \frac{\pi}{6}$For $n=0$,$	heta = \frac{\pi}{6}$ (Valid,$\cos \frac{\pi}{6} 
eq 0$)For $n=1$,$	heta = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$ (Outside range $[0, 2\pi]$)Case $2$: $	heta = 2n\pi - \frac{\pi}{3} - \frac{\pi}{6} = 2n\pi - \frac{\pi}{2}$For $n=1$,$	heta = 2\pi - \frac{\pi}{2} = \frac{3\pi}{2}$At $	heta = \frac{3\pi}{2}$,$\cos 	heta = 0$,so $\sec 	heta$ and $	an 	heta$ are undefined.Thus,the only valid solution in $[0, 2\pi]$ is $	heta = \frac{\pi}{6}$.The number of distinct solutions is $1$.

The number of distinct solutions of $\sec \theta + \tan \theta = \sqrt{3}$ for $0 \leqslant \theta \leqslant 2\pi$ is:

Similar Questions

The general solution of $\cos 2x - 2 \tan x + 2 = 0$ is

The set of values of $x$ satisfying the equation $2^{\tan(x - \pi/4)} - 2(0.25)^{\frac{\sin^2(x - \pi/4)}{\cos 2x}} + 1 = 0$ is:

If $x \in(-\pi, \pi)$ then the number of solutions of the equation $2 \sin x \sin 3 x \sin 5 x+\sin 5 x \cos 4 x=0$ is

The number of solutions of the equation $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0$ for $\theta \in [-3\pi, 2\pi]$ is:

The common principal solution of the equations $\sin \theta = -\frac{1}{2}$ and $\tan \theta = \frac{1}{\sqrt{3}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module