Question

Find the principal solutions of each of the following equations.

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

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

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

Answer 1

(i)The given equation is $sinx=\frac{1}{2}.$

We know that $sin\frac{\pi }{6}=\frac{1}{2}andsin(\pi -\frac{\pi }{6})=\frac{1}{2}.$

$\therefore sin\frac{\pi }{6}=\frac{1}{2}$ and $sin\frac{5\pi }{6}=\frac{1}{2}.$

Hence, the principal solutions are $x=\frac{\pi }{6}$ and $x=\frac{5x}{6}$.

(ii) The given equation is $cosx=\frac{1}{\sqrt{2}}.$

We know that $cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $cos(2\pi -\frac{\pi }{4})=\frac{1}{\sqrt{2}}.$

$\therefore cos\frac{\pi }{4}=\frac{1}{\sqrt{2}}$ and $cos\frac{7\pi }{4}=\frac{1}{\sqrt{2}}$

Hence, the principal solutions are $x=\frac{\pi }{4}$ and $x=\frac{7\pi }{4}.$

(iii)The given equation is $tanx=\frac{1}{\sqrt{3}}.$

We know that $tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}$ and $tan(\pi +\frac{\pi }{6})=\frac{1}{\sqrt{3}}.$

$\therefore tan\frac{\pi }{6}=\frac{1}{\sqrt{3}}$ and $tan\frac{7\pi }{6}=\frac{1}{\sqrt{3}}.$

Hence, the principal solutions are $x=\frac{\pi }{6}$ and $x=\frac{7\pi }{6}.$