Question

If $\sin {(x-20)}^{\circ }\cos {(3x-10)}^{\circ }=0$, Then find the value of x.

Answer 1

We have,

$\sin (x-{20}^o)\cos (3x-{10}^o)=0$

Then,

$\sin (x-{20}^o)\cos (x-{30}^o)=0$

$\Rightarrow \sin (x-{20}^o)=0,\cos (x-{20}^o)=0$

$\Rightarrow \sin (x-{20}^o)=\sin 0,\cos (x-{20}^o)=\cos \frac{\pi }{2}$

$\Rightarrow x-{20}^o=0,x-{20}^o=\frac{\pi }{2}$

$\Rightarrow x={20}^o,x-{20}^o={90}^o$

$\Rightarrow x={20}^o,x={70}^o$

Hence, this is the answer.