JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

The general solution of the equations:

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

and

\tan \theta=\sqrt{3}

View answer and explanation

Correct option: B

2 n \pi+\frac{4 \pi}{3}

Option B (

2 n \pi+\frac{{4 \pi}}{{3}}

) is the correct general solution because it satisfies both

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

and

\tan \theta=\sqrt{{3}}

More Questions from TRIGONOMETRIC EQUATIONS

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\mathrm{X}

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

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