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: 
$\frac{\sqrt{3}}{2}\cdot \cos x + \frac{1}{2}\sin x = \frac{1}{4}$

$\sin ({x + \frac{\pi}{3}}) = \frac{1}{4}$
$\sqrt3 cosx \ + \ sinx = \frac{1}{2}$ $\rightarrow$ $\ 2cos(30)cos(x) \ + 2sin(30)sin(x)\ = \frac{1}{2}$ $\rightarrow$ $2cos(x-30)$ = $\frac{1}{2}$ $\rightarrow$ $x-30$ = $arccos(\frac{1}{4})$ and from here you can deduce the angle and so the type of traingle.