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.
given that
$$\sqrt{3}cosx+sinx=\frac{1}{2}$$ Now, diving the above equation by $\sqrt{(\sqrt{3})^2+(1)^2}=2$ we get
$$\begin{align}
\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=\frac{1}{4}\\
cosxcos\frac{\pi}{6}+sinxsin\frac{\pi}{6}=\frac{1}{4}\\
cos\left(x-\frac{\pi}{6}\right)=\frac{1}{4}\\
x-\frac{\pi}{6}=cos^{-1}\frac{1}{4}\\
x=\frac{\pi}{6}+cos^{-1}\frac{1}{4}\\
\end{align}$$
$$x\approx1.841714847... rad\approx105.52^{o}>90^{o}$$
Thus, the triangle is an obtuse angled triangle. Hence, the option (a) is correct.