Find the principal solutions of each of the following equations :

To Find: Principal solution.

(i) Given: $\sin x=\frac{-1}{2}$

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

By using above formula, we have

$\sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin \left(\pi+\frac{\pi}{6}\right)=\sin \frac{7 \pi}{6} \Rightarrow x=n \pi+\frac{7 \pi}{6}(-1)^{n}$

Put $\mathrm{n}=0 \Rightarrow \mathrm{x}=0 \times \pi+\frac{7 \pi}{6}(-1)^{0} \Rightarrow \mathrm{x}=\frac{7 \pi}{6}$

Put $n=1 \Rightarrow x=1 \times \pi+\frac{7 \pi}{6}(-1)^{1} \Rightarrow x=1 \times \pi+\frac{7 \pi}{6}(-1)^{1} \Rightarrow x=\pi-\frac{7 \pi}{6}=-\frac{\pi}{6}$

[NOTE: $-\frac{\pi}{6}=\frac{11 \pi}{6}$ ]

So principal solution is $x=\frac{7 \pi}{6}$ and $\frac{11 \pi}{6}$

(ii) Given: $\sqrt{2} \cos x+1=0 \Rightarrow \cos x=\frac{-1}{\sqrt{2}}$

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

By using above formula, we have

$\cos x=\frac{-1}{\sqrt{2}}=\cos \frac{3 \pi}{4} \Rightarrow x=2 n \pi \pm \alpha, n \in i$

Put $n=0 \Rightarrow x=2 \times 0 \times \pi \pm \frac{3 \pi}{4} \Rightarrow x=\frac{3 \pi}{4}$

Put $n=1 \Rightarrow x=2 \pi \pm \frac{3 \pi}{4} \Rightarrow x=\frac{5 \pi}{4}, \frac{11 \pi}{4} \Rightarrow x=\frac{5 \pi}{4}, \frac{11 \pi}{4}$

$\left[\frac{11 \pi}{4}>2 \pi\right.$ So it is not include in principal solution $]$

So principal solution is $x=\frac{3 \pi}{4}$ and $\frac{5 \pi}{4}$

(iii) Given: $\tan x=-1$

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

By using above formula, we have

$\tan x=-1=\tan \frac{3 \pi}{4} \Rightarrow x=n \pi+\alpha, n \in i$

Put $n=0 \Rightarrow x=n \pi+\frac{3 \pi}{4} \Rightarrow x=\frac{3 \pi}{4}$

Put $n=1 \Rightarrow x=\pi+\frac{3 \pi}{4} \Rightarrow x=\frac{7 \pi}{4} \Rightarrow x=\frac{7 \pi}{4}$

So principal solution is $x=\frac{3 \pi}{4}$ and $\frac{7 \pi}{4}$

(iv) Given: $\sqrt{3} \operatorname{cosec} x+2=0 \Rightarrow \operatorname{cosec} x=\frac{-2}{\sqrt{3}}$

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

So $\sin x=\frac{-\sqrt{3}}{2}$

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

By using above formula, we have

$\sin x=\frac{-\sqrt{3}}{2}=\sin \frac{4 \pi}{3} \Rightarrow \theta=n \pi+\frac{4 \pi}{3}(-1)^{n}$

Put $n=0 \Rightarrow x=0 \times \pi+\frac{4 \pi}{3}(-1)^{0} \Rightarrow x=\frac{4 \pi}{3}$

Put $n=1 \Longrightarrow x=1 \times \pi+\frac{4 \pi}{3}(-1)^{1} \Rightarrow x=1 \times \pi+\frac{4 \pi}{3}(-1)^{1} \Rightarrow x=\pi-\frac{4 \pi}{3}=\frac{-\pi}{3}$

[NOTE: $\frac{-\pi}{3}=\frac{5 \pi}{3}$ ]

So principal solution is $x=\frac{4 \pi}{3}$ and $\frac{5 \pi}{3}$

(v) Given: $\tan x=-\sqrt{3}$

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

By using above formula, we have

$\tan x=-\sqrt{3}=\tan \frac{2 \pi}{3} \Rightarrow x=n \pi+\alpha, n \in i$

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

Put $n=1 \Rightarrow x=\pi+\frac{2 \pi}{3} \Rightarrow x=\frac{5 \pi}{3}$

So principal solution is $x=\frac{2 \pi}{3}$ and $\frac{5 \pi}{3}$

(vi) Given: $\sqrt{3} \sec x+2=0 \Rightarrow \sec x=\frac{-2}{\sqrt{3}}$

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

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

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

By using the above formula, we have

$\cos x=\frac{\sqrt{3}}{2}=\cos \frac{5 \pi}{6} \Rightarrow x=2 n \pi \pm \alpha, n \in I$

Put $n=0 \Rightarrow x=2 n \pi \pm \frac{5 \pi}{6} \Rightarrow x=\frac{5 \pi}{6}$

Put $n=1 \Rightarrow x=2 \pi \pm \frac{5 \pi}{6} \Rightarrow x=\frac{7 \pi}{6}, \frac{17 \pi}{6} \Rightarrow x=\frac{7 \pi}{6}, \frac{17 \pi}{6}$

$\left[\frac{17 \pi}{6}>2 \pi\right.$ So it is not include in principal solution $]$

So principal solution is $x=\frac{5 \pi}{6}$ and $\frac{7 \pi}{6}$