The principal solutions of {cosec}(θ)=-2 are…

Correct Answer is : 7π6,11π67 π/6, 11 π/667π,611π

Correct Answer is : 7π6,11π67 π/6, 11 π/667π,611π

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MHTCET 11 Apr 2026 Shift 1 Paper

Question : 105 of 150

Marks: +1, -0

\operatorname{cosec}(\theta)=-2

are…

Solution:

Concept:

Use

\operatorname{cosec} \theta = \frac{1}{\sin \theta}

and find angles in

[0, 2\pi)

where

\sin \theta

is negative.

Explanation:

Given:

\operatorname{cosec} \theta = -2

Since

\operatorname{cosec} \theta = \frac{1}{\sin \theta}

, we get

\sin \theta = -\frac{1}{2}

\sin \theta

is negative in the third and fourth quadrants.

The reference angle for

\sin \theta = \frac{1}{2}

\frac{\pi}{6}

In the third quadrant:

\theta = \pi + \frac{\pi}{6} = \frac{7\pi}{6}

In the fourth quadrant:

\theta = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}

Thus, the principal solutions are

\frac{7\pi}{6}

and

\frac{11\pi}{6}

Answer:

Option C:

\frac{7\pi}{6}, \frac{11\pi}{6}

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