Solve $\sqrt{3}\sin(x)+\cos(x)-2=0$

Question

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?

Answer 1

An idea:

$$\sqrt3\,\sin x+\cos x-2=0\implies\frac{\sqrt3}2\sin x+\frac12\cos x=1\iff\sin\left(x+\frac\pi6\right)=1\implies$$

$$\implies x+\frac\pi6=\frac\pi2+2k\pi\implies x=\frac\pi3+2k\pi$$

Answer 2

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 $k\in\mathbb Z$.

Answer 3

Your mistake was forgetting that when you square a non-$0$ equality, the result satisfies two possible equations. In particular, both $x = y$ and $x = -y$ gives that $x^2 = y^2$. From what you describe, you did the following:

$$\sqrt{3}\sin(x) = 2 - \cos(x) \tag{1}\label{eq1}$$

then square both sides to get

$$3\sin^2(x) = 4 - 4\cos(x) + \cos^2(x) \tag{2}\label{eq2}$$

Next, you made certain manipulations to get

$3 - 3\cos^2(x) = 4 - 4\cos(x) + \cos^2(x) \; \Rightarrow \; 4\cos^2(x) - 4\cos(x) + 1 = 0$

Using $\cos(x) = t$ then gives

$$4t^2 - 4t + 1 = 0 \; \Rightarrow \; (2t - 1)^2 = 0 \; \Rightarrow \; t = \frac{1}{2} \tag{3}\label{eq3}$$

The full set of solutions for $\cos(x) = \frac{1}{2}$ is $x = \pm\frac{\pi}{3} + 2k\pi, k \in \mathbb{Z}$. However, note the RHS of \eqref{eq1} is always $\frac{3}{2}$, but for $x = \frac{\pi}{3} + 2k\pi$ the LHS is also $\frac{3}{2}$, but for $x = -\frac{\pi}{3} + 2k\pi$ the LHS is $-\frac{3}{2}$ instead. Squaring with either case still gives \eqref{eq2}.

Whenever you use a non-reversible operation, like squaring, it's important you check to remove any extraneous results you may have got from your manipulations. However, checking by substituting your results into the original equation is generally always a good idea in case you made some mistake.