Find the principal solutions of each of the following equations:

Solution:

To Find: Principal solution.

[NOTE: The solutions of a trigonometry equation for which 0 ≤ x < 2 ???? is called principal solution]

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

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

By using above formula, we have

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

Put $n=0 \Rightarrow x=0 \times \pi+\frac{\pi}{3}(-1)^{0} \Rightarrow x=\frac{\pi}{3}$

Put $n=1 \Rightarrow x=1 \times \pi+\frac{\pi}{3}(-1)^{1} \Rightarrow x=1 \times \pi+\frac{\pi}{3}(-1)^{1} \Rightarrow x=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

So principal solution is $x=\frac{\pi}{3}$ and $\frac{2 \pi}{3}$

(ii) Given: $\cos x=\frac{1}{2}$

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

By using above formula, we have

$\cos x=\frac{1}{2}=\cos \frac{\pi}{3} \Rightarrow \theta=2 n \pi \pm \alpha, n \in 1$

Put $n=0 \Rightarrow x=2 n \pi \pm \frac{\pi}{3} \Rightarrow x=\frac{\pi}{3}$

Put $n=1 \Longrightarrow x=2 \pi \pm \frac{\pi}{3} \Rightarrow x=\frac{5 \pi}{3}, \frac{7 \pi}{3} \Rightarrow x=\frac{5 \pi}{3}, \frac{7 \pi}{3}$

$\left[\frac{7 \pi}{3}>2 \pi\right.$ So it is not include in principal solution $]$

So principal solution is $x=\frac{\pi}{3}$ and $\frac{5 \pi}{3}$

(iii) Given: $\tan \mathrm{x}=\sqrt{3}$

Formula used: $\tan \theta=\tan \alpha \Rightarrow \theta=n \pi \pm \alpha, n \in \mid$

By using above formula, we have

$\tan \mathrm{x}=\sqrt{3}=\tan \frac{\pi}{3} \Rightarrow \mathrm{x}=\mathrm{n} \pi+\alpha, \mathrm{n} \in 1$

Put $\mathrm{n}=0 \Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{\pi}{3} \Rightarrow \mathrm{x}=\frac{\pi}{3}$

Put $n=1 \Rightarrow x=\pi+\frac{\pi}{3} \Rightarrow x=\frac{4 \pi}{3} \Rightarrow x=\frac{4 \pi}{3}$

So principal solution is $x=\frac{\pi}{3}$ and $\frac{4 \pi}{3}$

We know that $\tan \theta \times \cot \theta=1$

So $\cot x=\sqrt{3} \Rightarrow \tan x=\frac{1}{\sqrt{3}}$

The formula used: $\tan \theta=\tan \alpha \Rightarrow \theta=n \pi \pm \alpha, n \in I$

By using the above formula, we have

$\tan x=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6} \Rightarrow \theta=n \pi+\alpha, n \in 1$

Put $n=0 \Longrightarrow x=n \pi+\frac{\pi}{6} \Longrightarrow x=\frac{\pi}{6}$

Put $n=1 \Longrightarrow x=\pi+\frac{\pi}{6} \Rightarrow x=\frac{7 \pi}{6}$

So principal solution is $x=\frac{\pi}{6}$ and $\frac{7 \pi}{6}$

(v) Given: $\operatorname{cosec} x=2$

We know that $\operatorname{cosec} \theta \times \sin \theta=1$

So $\sin x=\frac{1}{2}$

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

By using above formula, we have

$\sin x=\frac{1}{2}=\sin \frac{\pi}{6} \Rightarrow \theta=n \pi+\frac{\pi}{6}(-1)^{n}$

Put $n=0 \Longrightarrow \theta=0 \times \pi+\frac{\pi}{6}(-1)^{0} \Rightarrow \theta=\frac{\pi}{6}$

Put $\mathrm{n}=1 \Rightarrow \theta=1 \times \pi+\frac{\pi}{6}(-1)^{1} \Rightarrow \theta=1 \times \pi+\frac{\pi}{6}(-1)^{1} \Rightarrow \theta=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$

So principal solution is $x=\frac{\pi}{6}$ and $\frac{5 \pi}{6}$

(vi) Given: $\sec x=\frac{2}{\sqrt{3}}$

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

So $\cos x=\frac{\sqrt{3}}{2}$

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

By using the above formula, we have

$\cos x=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow x=2 n \pi \pm \alpha, n \in 1$

Put $n=0 \Longrightarrow x=2 n \pi \pm \frac{\pi}{6} \Rightarrow x=\frac{\pi}{6}$

Put $n=1 \Rightarrow x=2 \pi \pm \frac{\pi}{6} \Rightarrow x=\frac{11 \pi}{6}, \frac{13 \pi}{6} \Rightarrow x=\frac{11 \pi}{6}, \frac{13 \pi}{6}$

$\left[\frac{13 \pi}{6}>2 \pi\right.$ So it is not include in principal solution $]$

So principal solution is $x=\frac{\pi}{6}$ and $\frac{11 \pi}{6}$