Solving an elementary trigonometric equation

[brotherbobby]

Homework Statement

Solve $\boldsymbol{\sqrt{3}\cos x+\sin x = \sqrt 2}$

Relevant Equations

1. $\sin(A+B)=\sin A \cos B+\cos A\sin B$
2. If $\sin x = \sin\alpha\Rightarrow x=n\pi+(-1)^n\alpha$, where $n\in \mathbb{Z}$

Problem statement : I copy and paste the problem as it appeared in the text to the right.

Attempt :

$\quad \begin{matrix} \sqrt{3}\cos x +\sin x &=& \sqrt{2}\\ \Rightarrow\dfrac{\sqrt{3}}{2}\cos x + \dfrac{1}{2}\sin x &=& \dfrac{1}{\sqrt{2}}\\ \Rightarrow\sin\left(\dfrac{\pi}{3}+x\right)&=&\sin\dfrac{\pi}{4}\\ \Rightarrow\dfrac{\pi}{3}+x&=&n\pi+(-1)^n\dfrac{\pi}{4}\\ \Rightarrow x&=&n\pi+(-1)^n\dfrac{\pi}{4}-\dfrac{\pi}{3}\\ \end{matrix}$
So the answer is $\boxed{x=n\pi+(-1)^n\dfrac{\pi}{4}-\dfrac{\pi}{3}}$ , where $n\in\mathbb Z$.

Taking (non-negative) values of $n=0,1,2,\dots$, we find $x=-\dfrac{\pi}{12}, \dfrac{11\pi}{12}, \dfrac{23\pi}{12}, \dots$

Doubt : My solution is correct, but there are a range of other answer that my method does not give. Briefly, that $x = 2n\pi+\dfrac{5\pi}{12}$.

Curiously, when I solve the equation as a "cosine", it comes to $\cos\left( x-\dfrac{\pi}{6}\right) = \cos\dfrac{\pi}{4}$, I get both the answers.

Request : Why doesn't my method, involving "sines", not work? Many thanks.

 

[haruspex]

So the answer is $\boxed{x=n\pi+(-1)^n\dfrac{\pi}{4}-\dfrac{\pi}{3}}$ , where $n\in\mathbb Z$.

I agree, but…

Taking (non-negative) values of $n=0,1,2,\dots$, we find $x=-\dfrac{\pi}{12}, \dfrac{11\pi}{12}, \dfrac{23\pi}{12}, \dots$

That’s not what I get at n=1.

 

[pasmith]

$$n\pi + (-1)^n \frac{\pi}4 - \frac{\pi}3 = n\pi + \frac{\pi}{12} (3(-1)^n - 4).$$ For even $n = 2m$ this is $2m\pi - \frac{\pi}{12}$. For odd $n = 2m+1$ it is
$$(2m+ 1)\pi - \frac{7\pi}{12} = 2m\pi + \frac{5\pi}{12}.$$

 

[wrobel]

Perhaps it would be clearer to write:
$$\sin x=a,\quad |a|\le 1\Longrightarrow$$
$$x=\arcsin a+2\pi n$$
and $$x=\pi-\arcsin a+2\pi k,\quad k,n\in\mathbb{Z}$$