Nature of the $\triangle$

Question

In $\triangle$ ABC, the $\angle BAC$ is a root of the equation $3^{1\over2} \cos x + \sin x = {1\over2}.$ Then what kind of triangle is the $\triangle$ ABC.

Answer 1

Multiply both sides of the equation by $1/2$ and use the expansion formula for $\sin(A+B)$ to get $$\sin(x+60^\circ)=\frac{1}{4}\Rightarrow x\approx 105.522^\circ$$ So the triangle is an obtuse triangle.

Answer 2

You can't know, because as it was said in comments, $$\sin(x+60°)=\frac{1}{4}$$

means $x\approx 105.522°$ or $x\approx -45.522°$. So either the triangle is obtuse or not.