Question 17.2.7: Find the general solution of each of the equations: (i) sin ......
Find the general solution of each of the equations:
(i) \sin 2 x=-\frac{1}{2} \quad(ii) \tan 3 x=-1
(i) \sin 2 x=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6} \\ \Rightarrow \sin 2 x=\sin \frac{7 \pi}{6} \\ \Rightarrow 2 x=\left\{n \pi+(-1)^{n} \cdot \frac{7 \pi}{6}\right\} , where n \in I \\ \Rightarrow x=\left\{\frac{n \pi}{2}+(-1)^{n} \cdot \frac{7 \pi}{12}\right\} , where n \in I .
Hence, the general solution is x=\left\{\frac{n \pi}{2}+(-1)^{n} \cdot \frac{7 \pi}{12}\right\} , where n \in I .
(ii) \tan 3 x=-1=-\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)=\tan \frac{3 \pi}{4} \\ \Rightarrow \tan 3 x=\tan \frac{3 \pi}{4} \\
\Rightarrow 3 x=\left(n \pi+\frac{3 \pi}{4}\right) , where n \in I .
\Rightarrow x=\left(\frac{n \pi}{3}+\frac{\pi}{4}\right) , where n \in I .
Hence, the general solution is x=\left(\frac{n \pi}{3}+\frac{\pi}{4}\right) , where n \in I .