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

Question:

Find the general solution of each of the following equations:

(i) $\sin x=\frac{\sqrt{3}}{2}$

(ii) $\cos x=1$

(iii) $\sec \mathrm{x}=\sqrt{2}$

Solution:

To Find: General solution.

(i) Given: $\sin x=\frac{\sqrt{3}}{2}$

Formula used: $\sin \theta=\sin \alpha \Rightarrow \theta=\mathrm{n} \pi+(-1)^{\mathrm{n}} \alpha, \mathrm{n} \in$ ।

By using above formula, we have

$\sin x=\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3} \Longrightarrow x=n \pi+(-1)^{n} \cdot \frac{\pi}{3}$

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

(ii) Given: $\cos x=1$

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

By using above formula, we have

$\cos x=1=\cos \left(0^{\circ}\right) \Rightarrow x=2 n \pi, n \in 1$

So general solution is $x=2 n \pi$ where $n \in I$

(iii) Given: $\sec x=\sqrt{2}$

We know that $\sec \theta \times \cos \theta=1$

So $\cos x=\frac{1}{\sqrt{2}}$

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

So general solution is $x=2 n \pi \pm \frac{\pi}{4}$ where $n \in I$