Question

Find the principal and general solutions of $\mathrm{cosec}x=-2$

Answer 1

Step 1: Define the problem

Given, $\mathrm{cosec}x=-2$

$$\begin{gathered} \Rightarrow \frac{1}{\sin \left(x\right)}=-2 \\ \Rightarrow \sin \left(x\right)=-\frac{1}{2} \end{gathered}$$

Step 2: Find the value of $x$ in $3rd$quadrant

We know that $\sin (30°)=\frac{1}{2}$

Since, $\sin (180°+\alpha )=-\sin \left(\alpha \right)$ for some angle $\alpha$.

Thus, $\sin (180+30)=-\sin \left(30\right)\Rightarrow \sin \left(210\right)=-\frac{1}{2}$

Converting from degrees to radians,

$$\begin{gathered} \sin \left(210·\frac{\mathrm{\pi }}{180}\right)=-\frac{1}{2} \\ \Rightarrow \sin \left(\frac{7\mathrm{\pi }}{6}\right)=-\frac{1}{2} \\ \Rightarrow \mathrm{cosec}\left(\frac{7\mathrm{\pi }}{6}\right)=-2 \end{gathered}$$

Therefore, $x=\frac{7\mathrm{\pi }}{6}$ is the principal value in the $3rd$quadrant

Step 3: Find the value of $x$ in $4th$ quadrant

Since, $\sin (360°-\alpha )=-\sin \left(\alpha \right)$ for some angle $\alpha$.

Thus, $\sin (360-30)=-\sin \left(30\right)\Rightarrow \sin \left(330\right)=-\frac{1}{2}$

Converting from degrees to radians,

$$\begin{gathered} \sin \left(330·\frac{\mathrm{\pi }}{180}\right)=-\frac{1}{2} \\ \Rightarrow \sin \left(\frac{11\mathrm{\pi }}{6}\right)=-\frac{1}{2} \\ \Rightarrow \mathrm{cosec}\left(\frac{11\mathrm{\pi }}{6}\right)=-2 \end{gathered}$$

Therefore, $x=\frac{11\mathrm{\pi }}{6}$ is the principal value in the$4th$ quadrant

Step 4: Find general solution

The values for $x$ satisfying the condition $\mathrm{cosec}\left(\mathrm{x}\right)=-2$ are $\frac{7\mathrm{\pi }}{6}$, $\frac{11\mathrm{\pi }}{6}$

These can be obtained from the general equation $x=n\mathrm{\pi }+{(-1)}^n\frac{7\mathrm{\pi }}{6},n\in \mathrm{\mathbb{Z}}$

Therefore, the principal values of $\mathrm{cosec}x=-2$ are $\frac{7\mathrm{\pi }}{6}$ and $\frac{11\mathrm{\pi }}{6}$and the general solution is $x=n\mathrm{\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6},n\in \mathrm{\mathbb{Z}}$.