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Question: Find the general solution of \(\csc x=-2\)....

Find the general solution of csc⁡x=−2\csc x=-2.

Explanation

Solution

Hint: First of all, we can write csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}. Now, we can find the principal solution of sin⁡x=−12\sin x=\dfrac{-1}{2}. By using the principal solution, we can find the general solution using sin⁡x=sin⁡α\sin x=\sin \alpha , then x=nπ+(−1)nαx=n\pi +{{\left( -1 \right)}^{n}}\alpha .

Complete step-by-step answer:
Here we have to find the general solution of csc⁡x=−2\csc x=-2. Let us consider the equation given in the question, csc⁡x=−2\csc x=-2. We know that csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}, so we get, 1sin⁡x=−2\dfrac{1}{\sin x}=-2. By cross multiplying the equation, we get, sin⁡x=−12\sin x=\dfrac{-1}{2}. We know that sin⁡30∘=12\sin 30{}^\circ =\dfrac{1}{2}. Since xx is negative, it will be in the third and fourth quadrants.

Value in the third quadrant =180∘+30∘=210∘=180{}^\circ +30{}^\circ =210{}^\circ . Value in the fourth quadrant =360∘−30∘=330∘=360{}^\circ -30{}^\circ =330{}^\circ . So, for sin⁡x=−12\sin x=\dfrac{-1}{2}, we get,
x=210∘=210×π180=7π6 x=330∘=330×π180=11π6 \begin{aligned} & x=210{}^\circ =210\times \dfrac{\pi }{180}=\dfrac{7\pi }{6} \\\ & x=330{}^\circ =330\times \dfrac{\pi }{180}=\dfrac{11\pi }{6} \\\ \end{aligned}
Now, to find the general solution, let sin⁡x=sin⁡θ………(i)\sin x=\sin \theta \ldots \ldots \ldots \left( i \right)
And we know that sin⁡x=−12………(ii)\sin x=\dfrac{-1}{2}\ldots \ldots \ldots \left( ii \right)
From equation (i) and equation (ii), we get,
sin⁡θ=−12………(iii)\sin \theta =\dfrac{-1}{2}\ldots \ldots \ldots \left( iii \right)
We have already calculated, sin⁡7π6=−12………(iv)\sin \dfrac{7\pi }{6}=\dfrac{-1}{2}\ldots \ldots \ldots \left( iv \right)
So, from the equations, (i), (iii) and (iv), we get,
sin⁡x=sin⁡θ=sin⁡7π6 ⇒sin⁡x=sin⁡7π6 \begin{aligned} & \sin x=\sin \theta =\sin \dfrac{7\pi }{6} \\\ & \Rightarrow \sin x=\sin \dfrac{7\pi }{6} \\\ \end{aligned}
We know that when sin⁡x=sin⁡θ\sin x=\sin \theta , then x=nπ+(−1)nθx=n\pi +{{\left( -1 \right)}^{n}}\theta . By using this relation, we get, x=nπ+(−1)n7π6x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{7\pi }{6}, when n∈Nn\in N.
Hence, we get the general solution of csc⁡x=−2\csc x=-2 as x=nπ+(−1)n7π6x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{7\pi }{6}.

Note: In these types of questions, instead of remembering all the multiple formulas for each trigonometric ratio, it is better to convert sec, cosec and cot into cos, sin and tan respectively and then solve the question. It is also advised to learn the values of sin, cos, and tan at the standard angles which are, 0∘,30∘,45∘,60∘,90∘0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ as we can always find the higher angles with the use of these angles.