Second solution to $ \sin(x) - \sqrt{2} \cos(x) = 1$? I think I'm missing a solution to the following problem:

Solutions to the equation $ \sin(x) - \sqrt{2} \cos(x) = 1$ are....

I get $x = 1.571 \pm 2 n \pi$. This matches the answer in the book I'm using, Engineering Mathematics, 5th Edition, by Stroud.
However, shouldn't there be 2 answers per period? Unlike their previous example, $ 4 \sin(x) - 3 \cos(x) = 5$, this solution isn't at the peak amplitude. I would think there is an additional solution as the function decreases after the peak?
I don't know how to find it if that's the case, or maybe I'm turned all around and thinking about this wrong. What am I missing? Thanks for your help.
 A: You are correct, there is another solution per period. I found it as follows. Start by squaring both sides to get
$$
(\sin(x)-\sqrt{2}\cos(x))^2 = \cos^2(x) - 2 \sqrt{2} \sin(x) \cos(x) + 1 = 1^2 = 1
$$
Then subtracting 1 from both sides, and dividing by $\cos^2(x)$ (you already found the solution where $\cos(x) = 0$) we have
$$
1 - 2 \sqrt{2} \tan(x) = 0.
$$
Solving then gives $\tan(x) = \sqrt{2}/4$, from which we find the solutions $\arctan(\sqrt{2}/4) + \pi n$. Checking these back in the original equation (we squared so we may have extra incorrect solutions!) we see that these solutions work exactly when $n$ is odd. That is, $x = \arctan(\sqrt{2}/4) + (2n+1)\pi$.
A: Let $\tan(x/2)=t$. then
$$\sin x-\sqrt{2} \cos x=1 \implies \frac{2t}{1+t^2}-\sqrt{2} \frac{1-t^2}{1+t^2}=1$$
$$\implies (\sqrt{2}-1)t^2+2t-(\sqrt{2}+1)=0 \implies t=1,-(3+2\sqrt{2})$$ So this equation has two branches of solutions:
$$x/2=\pi/4+n\pi \implies x=\pi/2+2n\pi,n0,1,2,...$$ and
$$x/2=-\tan^{-1}[(3+2\sqrt{2}]+m\pi\implies x=-2\tan^{-1}(3+2\sqrt{2})+2m\pi, m=0,1,2,..$$
$$\implies x\approx -0.89 \pi+2m\pi$$
In $[0,2\pi]$ two roots are $1.57, 3.48,$ when $n=0,m=1.$ See  the Fig. below:

A: $$\sin x-\sqrt2\cos x=\sqrt3\left(\frac1{\sqrt3}\sin x-\sqrt{\frac{2}3}\cos x\right)=\sqrt3\sin\left(x-\arccos\frac1{\sqrt3}\right)$$
Therefore $$\sin x-\sqrt2\cos x=1\iff\sin\left(x-\arccos\frac1{\sqrt3}\right)=\frac1{\sqrt3}$$
In $[0,2\pi)$, this has solutions $$x=\arcsin\frac1{\sqrt3}+\arccos \frac1{\sqrt3}\lor x=\pi-\arcsin\frac1{\sqrt3}+\arccos\frac1{\sqrt3}$$
which are $\frac\pi2$ and $\approx3.481$ respectively. So you're right, one is missing.
