# Find the general solution of each of the following equations:

Question:

Find the general solution of each of the following equations:

(i) $4 \cos ^{2} x=1$

(ii) $4 \sin ^{2} x-3=0$

(iii) $\tan ^{2} x=1$

Solution:

To Find: General solution.

(i) Given: $4 \cos ^{2} x=1 \Longrightarrow \cos ^{2} x=\left(\frac{1}{4}\right)$

$\therefore \cos ^{2} x=\cos ^{2} \frac{\pi}{3}$

Formula used: $\cos ^{2} \theta=\cos ^{2} \alpha \Rightarrow \theta=n \pi \pm \alpha, n \in I$

By using the above formula, we have

$x=n \pi \pm \frac{\pi}{3}, n \in I$

So the general solution is $x=n \pi \pm \frac{\pi}{3}$ where $n \in$ ।

(ii) Given: $4 \sin ^{2} x-3=0 \Rightarrow \sin ^{2} x=\frac{3}{4}=\sin ^{2} \frac{\pi}{3}$

$\therefore \sin ^{2} x=\sin ^{2} \frac{\pi}{3}$

Formula used: $\sin ^{2} \theta=\sin ^{2} \alpha \Rightarrow \theta=\mathrm{n} \pi \pm \alpha, \mathrm{n} \in I$

By using the above formula, we have

$x=n \pi \pm \frac{\pi}{3}, n \in I$

So the general solution is $x=n \pi \pm \frac{\pi}{3}$ where $n \in$ ।

(ii) Given: $\tan ^{2} x=1 \Rightarrow \tan ^{2} x=\tan ^{2} \frac{\pi}{4}$

$\therefore \tan ^{2} x=\tan ^{2} \frac{\pi}{4}$

The formula used: $\tan ^{2} \theta=\tan ^{2} \alpha \Rightarrow \theta=n \pi \pm \alpha, n \in$ ।

By using the above formula, we have

$x=n \pi \pm \frac{\pi}{4}, n \in I$

So the general solution is $x=n \pi \pm \frac{\pi}{4}$ where $n \in I$

So general solution is $x=\frac{n \pi}{3}+(-1)^{n} \cdot \frac{4 \pi}{9}$, where $n \in I$