Question

In each of the following, find the general value of x satisfying the equation:

$\left(i\right)sinx=\frac{1}{\sqrt{2}}$

$\left(ii\right)cosx=\frac{1}{2}$

$\left(iii\right)tanx=\frac{1}{\sqrt{3}}$

Answer 1

(i)Given: $sinx=\frac{1}{\sqrt{2}}.$

The least value of x in $[0,2\pi )$
$\text{[for which}$ $sinx=\frac{1}{\sqrt{2}}$ $is$ $x=\frac{\pi }{4}.$

$\therefore sinx=sin\frac{\pi }{4}\Rightarrow x=n\pi +(-1{)}^n.\frac{\pi }{4}$, where $n\in I.$

Hence, the general solution is $x=n\pi +(-1{)}^n.\frac{\pi }{4}$, where $n\in I.$

(ii)Given: $cosx=\frac{1}{2}.$

The least value of x in $[0,2\pi )$
$\text{[for which}$ $cosx+\frac{1}{2}$ $is$ $x=\frac{\pi }{3}.$

$\therefore cosx=cos\frac{\pi }{3}\Rightarrow x=(2n\pi \pm \frac{\pi }{3})$, where $n\in I.$

Hence, the general solution is $x=(2n\pi \pm \frac{\pi }{3})$, where $n\in I.$