I need to solve the equation $$\sqrt{3}\sin(x)+\cos(x)-2=0$$
My try: I separated the radical then I squared and I noted $\cos(x)=t$ and I got a quadratic equation with $t=\frac{1}{2}$
To solve $\cos(x)=\frac{1}{2}$ I used formula $x\in\left \{ +-\arccos(\frac{1}{2})+2k\pi \right \}k\in\mathbb{Z}$ and I got $x\in\left \{ -+ \frac{\pi}{3}+2k\pi \right \}$ but the right answer is $x=(-1)^{k}\cdot\frac{\pi}{2}+k\pi-\frac{\pi}{6}$
Where's my mistake?How to get the right answer?
Best Answer
I like the following way.
By C-S $$2=\sqrt3\sin{x}+\cos{x}\leq\sqrt{((\sqrt3)^2+1^2)(\sin^2x+\cos^2x)}=2.$$ The equality occurs for $$(\sqrt3,1)||(\sin{x},\cos{x})$$ or since $$\sqrt3\sin{x}+\cos{x}\geq0,$$ for $$\sin{x}=\frac{\sqrt3}{2}$$ and $$\cos{x}=\frac{1}{2},$$ which gives $$x=60^{\circ}+360^{\circ}k,$$ where $k\in\mathbb Z$.