The number of solutions of the equation tan t | vedclass

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

Question

(B) Given trigonometric equation is,$	an 	heta + \sec 	heta = 2 \cos 	heta$. Multiplying by $\cos 	heta$ (since $\cos 	heta 
eq 0$): $\frac{\si

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given trigonometric equation is,$	an 	heta + \sec 	heta = 2 \cos 	heta$. Multiplying by $\cos 	heta$ (since $\cos 	heta 
eq 0$): $\frac{\sin 	heta}{\cos 	heta} + \frac{1}{\cos 	heta} = 2 \cos 	heta$ $\Rightarrow 1 + \sin 	heta = 2 \cos^2 	heta$ Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$: $1 + \sin 	heta = 2(1 - \sin^2 	heta)$ $2 \sin^2 	heta + \sin 	heta - 1 = 0$ Factoring the quadratic: $(2 \sin 	heta - 1)(\sin 	heta + 1) = 0$ This gives $\sin 	heta = \frac{1}{2}$ or $\sin 	heta = -1$. For $\sin 	heta = \frac{1}{2}$ in $(0, 2 \pi)$,$	heta = \frac{\pi}{6}$ or $	heta = \frac{5 \pi}{6}$. For $\sin 	heta = -1$ in $(0, 2 \pi)$,$	heta = \frac{3 \pi}{2}$. However,the original equation involves $	an 	heta$ and $\sec 	heta$,which are undefined at $	heta = \frac{3 \pi}{2}$ (since $\cos \frac{3 \pi}{2} = 0$). Thus,the valid solutions are $	heta = \frac{\pi}{6}$ and $	heta = \frac{5 \pi}{6}$. Therefore,the number of solutions is $2$.

The number of solutions of the equation $\tan \theta + \sec \theta = 2 \cos \theta$ where $\cos \theta \neq 0$ lying in the interval $(0, 2 \pi)$ is

Similar Questions

The number of solutions of $\sin^2 \theta = \frac{1}{2}$ in the interval $[0, \pi]$ is . . . . . .

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

The number of solutions of the equation $\sin A - 5 \sin 2A + \sin 3A = \cos A - 5 \cos 2A + \cos 3A$ in the interval $(0, \pi)$ is:

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

If $\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = 3$,then the general value of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module