Total number of solutions of $\sin 2 x-\sin x-1=0$ in $[-2 \pi, 2 \pi]$ is equal to

JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Total number of solutions of

\sin 2 x-\sin x-1=0

[-2 \pi, 2 \pi]

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct. By simplifying the equation to

\sin x = \frac{{1 - \sqrt{{5}}}}{{2}}

, we find that there are 4 solutions for

x

in the interval

[-2\pi, 2\pi]

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