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The value of $θ (0 ° < θ < 360 °)$ satisfying $\cos e c θ + 2 = 0$ are

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210 °, 240 °

210 °, 300 °

240 °, 300 °

210 °, 330 °

answer is D.

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Detailed Solution

\cos e c θ = - 2

$\sin θ = - \frac{1}{2}$

$θ \in Q_{3}, Q_{4}$

$θ = 180 ° + 30 °, 360 ° - 30 °$

$= 210 °, 330 °$

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