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Question: Value of \( \theta (0 < \theta < 360^\circ ) \) which satisfy the equation \( \csc \theta + 2 = 0 \)...

Value of θ(0<θ<360∘)\theta (0 < \theta < 360^\circ ) which satisfy the equation csc⁡θ+2=0\csc \theta + 2 = 0
(A) 210∘,100∘210^\circ ,100^\circ
(B) 240∘,300∘240^\circ ,300^\circ
(C) 210∘,240∘210^\circ ,240^\circ
(D) 210∘,330∘210^\circ ,330^\circ

Explanation

Solution

Hint : Simplify the equation csc⁡θ+2=0\csc \theta + 2 = 0 to get sin⁡θ=−12\sin \theta = - \dfrac{1}{2} . Use the fact that sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2} and sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x . Compute sin⁡(180+30)∘\sin (180 + 30)^\circ and sin⁡(360−30)∘\sin (360 - 30)^\circ to get the answer.

Complete step-by-step answer :
We are given the equation csc⁡θ+2=0\csc \theta + 2 = 0 .
We need to find a value of θ\theta such that csc⁡θ+2=0\csc \theta + 2 = 0 and 0<θ<360∘0 < \theta < 360^\circ .
Consider the equation csc⁡θ+2=0\csc \theta + 2 = 0 .
Then csc⁡θ=−2.......(1)\csc \theta = - 2.......(1)
We know that sin⁡θ=1csc⁡θ\sin \theta = \dfrac{1}{{\csc \theta }} .
Taking reciprocal on both sides, we get
1csc⁡θ=1−2 ⇒sin⁡θ=−12  \dfrac{1}{{\csc \theta }} = \dfrac{1}{{ - 2}} \\\ \Rightarrow \sin \theta = - \dfrac{1}{2} \\\
Here sin⁡θ\sin \theta is negative, therefore, θ\theta will be in the third or fourth quadrant.

We know that sin⁡30∘=12\sin 30^\circ = \dfrac{1}{2} , sin⁡(−x)=−sin⁡x\sin ( - x) = - \sin x for any 0<x<360∘0 < x < 360^\circ .
Therefore, sin⁡(−30)∘=−12\sin ( - 30)^\circ = - \dfrac{1}{2}
We know that sine is a periodic function with its period being 360∘360^\circ or 2π2\pi .
Therefore, sin⁡(−30+n360)∘=−12\sin ( - 30 + n360)^\circ = - \dfrac{1}{2} for any integer n but we have the condition that 0<θ<360∘0 < \theta < 360^\circ
Therefore n = 1, and sin⁡(360−30)∘=−12=sin⁡330∘\sin (360 - 30)^\circ = - \dfrac{1}{2} = \sin 330^\circ .
Also, sin⁡(180+x)=−sin⁡x\sin (180 + x) = - \sin x for any 0<x<360∘0 < x < 360^\circ .
Therefore, sin⁡(180+30)∘=−sin⁡30∘=−12=sin⁡210∘\sin (180 + 30)^\circ = - \sin 30^\circ = - \dfrac{1}{2} = \sin 210^\circ
Hence the value of θ\theta is 210∘,330∘210^\circ ,330^\circ .
So, the correct answer is “Option D”.

Note : Identifying the quadrants where the value of θ\theta lies is one of the crucial steps to solving such questions. The value of sin⁡θ\sin \theta is positive in the first and second quadrants and negative in the third and fourth quadrants. The value of cos⁡θ\cos \theta is positive in the first and the fourth quadrants and negative in the remaining ones.