<s>[INST] Find the type of triangle from equation. 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.
 [/INST] 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. 
</s>