If frac{1}{6} sin theta, cos theta, tan theta | vedclass

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

Question

(A) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $G.P.$,we have $\cos^2 	heta = \frac{1}{6} \sin 	heta \cdot 	an 	heta$.

Vedclass · Accepted Answer

$2 n \pi \pm \frac{\pi}{3}$

Vedclass · Answer

(A) Given that $\frac{1}{6} \sin 	heta, \cos 	heta, 	an 	heta$ are in $G.P.$,we have $\cos^2 	heta = \frac{1}{6} \sin 	heta \cdot 	an 	heta$. Substituting $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we get $\cos^2 	heta = \frac{\sin^2 	heta}{6 \cos 	heta}$. This implies $6 \cos^3 	heta = \sin^2 	heta = 1 - \cos^2 	heta$. Rearranging gives $6 \cos^3 	heta + \cos^2 	heta - 1 = 0$. Let $x = \cos 	heta$. Then $6x^3 + x^2 - 1 = 0$. Testing $x = \frac{1}{2}$,we get $6(\frac{1}{8}) + \frac{1}{4} - 1 = \frac{3}{4} + \frac{1}{4} - 1 = 0$. Thus,$(2 \cos 	heta - 1)$ is a factor. Dividing $6x^3 + x^2 - 1$ by $(2x - 1)$ gives $3x^2 + 2x + 1 = 0$. The discriminant of $3x^2 + 2x + 1$ is $D = 2^2 - 4(3)(1) = 4 - 12 = -8 Therefore,$\cos 	heta = \frac{1}{2}$,which implies $	heta = 2n\pi \pm \frac{\pi}{3}$.

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

Similar Questions

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 number of solutions that the equation $\sin 5\theta \cos 3\theta = \sin 9\theta \cos 7\theta$ has in the interval $[0, \frac{\pi}{4}]$ is

The general solution of $\sin x + \cos x = 1$ is

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module