Question

Find the general solutions of the following equations:
(i) $\sin x=\frac{1}{2}$

(ii) $\cos x=-\frac{\sqrt{3}}{2}$

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

(iv) $\sec x=\sqrt{2}$

(v) $\tan x=-\frac{1}{\sqrt{3}}$

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

Answer 1

We have:
(i) $\sin x=\frac{1}{2}$
The value of x satisfying $\sin x=\frac{1}{2}$ is $\frac{\mathrm{\pi }}{6}$.
∴ $\sin x=\frac{1}{2}$
⇒ $\sin x=\sin \frac{\mathrm{\pi }}{6}$
⇒ $x=n\mathrm{\pi }+(-1{)}^n\frac{\mathrm{\pi }}{6}$, $n\in Z$

(ii) $\cos x=-\frac{\sqrt{3}}{2}$
The value of x satisfying $\cos x=-\frac{\sqrt{3}}{2}$ is $\frac{7\mathrm{\pi }}{6}$.
∴ $\cos x=-\frac{\sqrt{3}}{2}$
⇒ $\cos x=\cos \frac{7\mathrm{\pi }}{6}$
⇒ $x=2n\mathrm{\pi }\pm \frac{7\mathrm{\pi }}{6}$, $n\in Z$

(iii) $\mathrm{cosec}x=-\sqrt{2}$(or) $\sin x=-\frac{1}{\sqrt{2}}$
The value of x satisfying $\sin x=-\frac{1}{\sqrt{2}}$ is $-\frac{\mathrm{\pi }}{4}$.
∴ $\sin x=-\frac{1}{\sqrt{2}}$
⇒ $\sin x=\sin (-\frac{\mathrm{\pi }}{4})$
⇒ $x=n\mathrm{\pi }+{\left(-1\right)}^n\left(-\frac{\mathrm{\pi }}{4}\right)$, $n\in Z$
⇒ $x=n\mathrm{\pi }+(-1{)}^{n+1}\frac{\mathrm{\pi }}{4},\mathrm{n}\in Z$

(iv) $\sec x=\sqrt{2}$(or) $\cos x=\frac{1}{\sqrt{2}}$
The value of x satisfying $\cos x=\frac{1}{\sqrt{2}}$ is $\frac{\mathrm{\pi }}{4}$.
∴ $\cos x=\frac{1}{\sqrt{2}}$
⇒ $\cos x=\cos \frac{\pi }{4}$
⇒ $x=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}$, $n\in Z$

(v) $\tan x=-\frac{1}{\sqrt{3}}$
The value of x satisfying $\tan x=-\frac{1}{\sqrt{3}}$ is $-\frac{\mathrm{\pi }}{6}$.
∴ $\tan x=-\frac{1}{\sqrt{3}}$
⇒ $\tan x=\tan (-\frac{\mathrm{\pi }}{6})$
⇒ $x=n\mathrm{\pi }-\frac{\mathrm{\pi }}{6}$, $n\in Z$

(vi) $\sqrt{3}\sec x=2$
⇒ $\sec x=\frac{2}{\sqrt{3}}$ (or) $\cos x=\frac{\sqrt{3}}{2}$
The value of x satisfying $\cos x=\frac{\sqrt{3}}{2}$ is $\frac{\mathrm{\pi }}{6}$.
∴ $\cos x=\frac{\sqrt{3}}{2}$
⇒ $\cos x=\cos \frac{\mathrm{\pi }}{6}$
⇒ $x=2n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{6}$, $n\in Z$