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.

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.