Question

If $0^o\le x\le +{90}^o$ and $\sin x+\sqrt{3}\cos x=1$ then what is the value of $x$?

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

The correct option is A ${30}^o$

$sinx+\sqrt{3}cosx=1,\theta \le x\le +{90}^{\circ }$

Dividing both sides with 2

$\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx=1/2$

$cos{60}^{\circ }sinx+sin{60}^{\circ }cosx=1/2$

$sin({60}^{\circ }+x)=1/2$

$\therefore {60}^{\circ }+x=si{n}^{-1}\left(1/2\right)$

$\Rightarrow {60}^{\circ }+|x|={30}^{\circ }$

$\Rightarrow |x|=-{30}^{\circ }$

$(x={30}^{\circ })$

Option A is correct.