Question 17.2.5: In each of the following, find the general value of x satisf......
In each of the following, find the general value of x satisfying the equation:
(i) \sin x=\frac{-\sqrt{3}}{2} \quad(ii) \cos x=\frac{-1}{2}\quad (iii) \cot x=-\sqrt{3}
(i) \sin x=\frac{-\sqrt{3}}{2}=-\sin \frac{\pi}{3}=\sin \left(\pi+\frac{\pi}{3}\right)=\sin \frac{4 \pi}{3} .
\therefore \sin x=\sin \frac{4 \pi}{3} \Rightarrow x=\left\{n \pi+(-1)^{n} \cdot \frac{4 \pi}{3}\right) , where n \in I .
Hence, the general solution is x=\left\{n \pi+(-1)^{n} \cdot \frac{4 \pi}{3}\right) , where n \in I .
(ii) \cos x=\frac{-1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3} .
\therefore \cos x=\cos \frac{2 \pi}{3} \Rightarrow x=\left(2 n \pi \pm \frac{2 \pi}{3}\right) , where n \in I .
Hence, the general solution is x=\left(2 n \pi \pm \frac{2 \pi}{3}\right) , where n \in I .
(iii) \cot x=-\sqrt{3} \\\begin{array}{l}\Rightarrow \tan x=\frac{-1}{\sqrt{3}}=-\tan \frac{\pi}{6}=\tan \left(\pi-\frac{\pi}{6}\right)=\tan \frac{5 \pi}{6}\\ \\\Rightarrow \tan x=\tan \frac{5 \pi}{6}\\ \\\Rightarrow x=\left(n \pi+\frac{5 \pi}{6}\right), \text { where } n \in I .\end{array}
Hence, the general solution is x=\left(n \pi+\frac{5 \pi}{6}\right) , where n \in I .