Question 17.2.8: Find the general solution of each of the equations: (i) 4sin......
Find the general solution of each of the equations:
(i) 4 \sin ^{2} x=1 \quad (ii) 2 \cos ^{2} x=1 \quad(iii) \cot ^{2} x=3
(i) 4 \sin ^{2} x=1 \Rightarrow \sin ^{2} x=\frac{1}{4}=\left(\frac{1}{2}\right)^{2}=\sin ^{2} \frac{\pi}{6} \\ \Rightarrow \sin ^{2} x=\sin ^{2} \frac{\pi}{6} \\ \Rightarrow x=\left\{n \pi \pm \frac{\pi}{6}\right\} , where n \in I .
Hence, the general solution is x=\left(n \pi \pm \frac{\pi}{6}\right), n \in I .
(ii) 2 \cos ^{2} x=1 \Rightarrow \cos ^{2} x=\frac{1}{2}=\left(\frac{1}{\sqrt{2}}\right)^{2}=\cos ^{2} \frac{\pi}{4} \\\begin{array}{l}\Rightarrow \cos ^{2} x=\cos ^{2} \frac{\pi}{4} \\\Rightarrow x=\left(n \pi \pm \frac{\pi}{4}\right) \text {, where } n \in I .\end{array}
Hence, the general solution is x=\left(n \pi \pm \frac{\pi}{4}\right), n \in I .
(iii) \cot ^{2} x=3 \Rightarrow \tan ^{2} x=\frac{1}{3}=\left(\frac{1}{\sqrt{3}}\right)^{2}=\tan ^{2} \frac{\pi}{6} \\\begin{array}{l}\Rightarrow \tan ^{2} x=\tan ^{2} \frac{\pi}{6} \\\Rightarrow x=\left(n \pi \pm \frac{\pi}{6}\right), \text { where } n \in I .\end{array}
Hence, the general solution is \left(n \pi \pm \frac{\pi}{6}\right) , where n \in I .