If $1 + \sin \,\,\theta + {\sin ^2}\,\,\theta + \ldots $ to $\infty = 4 + 2\,\sqrt 3 ,\,\,0 < \theta < \pi ,\,\,\theta \ne \frac{\pi }{2}$, then $\theta = $

$\frac{\pi }{3}\;\,\,{\rm{or}},\,\frac{{2\;\pi }}{3}$

$\frac{\pi }{3}\;\,\,{\rm{or}},\,\,\frac{\pi }{6}$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

1 + \sin \,\,\theta + {\sin ^2}\,\,\theta + \ldots

\infty = 4 + 2\,\sqrt 3 ,\,\,0 < \theta < \pi ,\,\,\theta \ne \frac{\pi }{2}

, then

\theta =

View answer and explanation

Correct option: D

\frac{\pi }{3}\;\,\,\mathrm{or},\,\frac{{2\;\pi }}{3}

Option D (

\frac{\pi}{3}

\frac{2\pi}{3}

) is correct because solving the infinite geometric series equation results in

\sin \theta = \frac{\sqrt{3}}{2}

, which gives the solutions

\theta = \frac{\pi}{3}

and

\theta = \frac{2\pi}{3}

Practice All Chapter MCQs →

\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x

\left[ {0,\;\,\frac{\pi }{2}} \right]

\left( {x,\,y} \right)

x,\,\,y \in \left[ {0,10} \right]

\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1

\mathrm{X}

\theta \rightarrow \frac{\pi}{2}

\sin x+\cos x=\frac{\sqrt{7}}{2}

x \in\left[0, \frac{\pi}{4}\right]

\tan \frac{x}{2}

\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}

0 < x < \pi

\cos x+\sin x=\frac{1}{2}

\tan x