Question

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

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

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

(iv) $\sec \mathrm{\theta }=\sqrt{2}$

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

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

Answer 1

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

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

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

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

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

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