Question: In triangle $ABC$, the angle($BAC$) is a root of the equation 

$$\sqrt{3}\cos x + \sin x = \frac{1}{2}.$$ 

Then the triangle $ABC$ is

a) obtuse angled 

b) right angled

c) acute angled but not equilateral

d) equilateral.

Thanks in advance.

Answer: 
We have
$$\dfrac{\sqrt3}2 \cos(x) + \dfrac12 \sin(x) = \dfrac14 \implies \sin(x+\pi/3) = \dfrac14$$
We hence obtain $x+\pi/3 = \arcsin(1/4)$ or $x+\pi/3 = \pi - \arcsin(1/4)$
Note that $\arcsin(1/4) < \arcsin(1/2) = \dfrac{\pi}6 < \dfrac{\pi}3$. Since $x \in (0,\pi)$, we have
$$x+\pi/3 = \pi - \arcsin(1/4) \implies x = \pi - \left(\arcsin(1/4)+\pi/3\right) > \pi/2$$
Hence, the triangle is obtuse.
$\sqrt3 cosx \ + \ sinx = \frac{1}{2}$ $\rightarrow$ $\ 2cos(30)cos(x) \ + 2sin(30)sin(x)\ = \frac{1}{2}$ $\rightarrow$ $2cos(x-30)$ = $\frac{1}{2}$ $\rightarrow$ $x-30$ = $arccos(\frac{1}{4})$ and from here you can deduce the angle and so the type of traingle.