The most general value of ${\rm{\theta }}$ satisfying the equation ${(1 + 2\sin {\rm{\theta }})^2} + \,\,{(\sqrt 3 \tan {\rm{\theta }} - 1)^2} = 0$ are given by

$n\pi + {\left( { - 1} \right)^n}\frac{{7\pi }}{6}$

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The most general value of

\mathrm{\theta }

satisfying the equation

{(1 + 2\sin \mathrm{\theta })^2} + \,\,{(\sqrt 3 \tan \mathrm{\theta } - 1)^2} = 0

are given by

View answer and explanation

Correct option: C

2n\pi + \frac{{7\pi }}{6}

Option C (

2n\pi + \frac{{7\pi }}{6}

) is correct because it satisfies both equations derived from the given condition.

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