Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $
Hint: I think the approach of @JanEerland is instructive. Here some thoughts how we could find this kind of substitution.
When looking at \begin{align*} \sin x+\cos x=\frac{1}{3} \end{align*} and we think of the trigonometric addition formulas we know that \begin{align*} \sin(x+a)=\sin x \cos a+\cos x\sin a \end{align*} It would be convenient if $\cos a=\sin a$, so that we can separate them. This is the case if $a=\frac{\pi}{4}$ and we obtain \begin{align*} \sin\left(x+\frac{\pi}{4}\right) &=\sin x\sin \frac{\pi}{4}+\cos x\cos \frac{\pi}{4}\\ &=\frac{\sqrt{2}}{2}\left(\sin x+\cos x\right) \end{align*} It follows \begin{align*} \sin x+\cos x&=\frac{1}{3}\\ \sin\left(x+\frac{\pi}{4}\right)&=\frac{1}{3\sqrt{2}} \end{align*}
$$\cos(x)+\sin(x)=\frac{1}{3}\Longleftrightarrow$$
Use:
$$\cos(x)+\sin(x)=\sqrt{2}\left[\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right]=$$ $$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)$$
$$\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{3}\Longleftrightarrow$$ $$\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{3\sqrt{2}}$$
Now, when we take the inverse sine of both sides, we got two options, with $n_1\space\wedge\space n_2\in\mathbb{Z}$:
- $$\frac{\pi}{4}+x=\pi-\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow x=\frac{3\pi}{4}-\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1$$
- $$\frac{\pi}{4}+x=\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow x=\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_2-\frac{\pi}{4}$$
we have after squaring $$\sin(x)^2+\cos(x)^2+\sin(2x)=\frac{1}{9}$$ or $$\sin(2x)=-\frac{8}{9}$$ the answer is $$c_1\in \mathbb{Z}\land \left(x=2 \pi c_1+2 \tan ^{-1}\left(\frac{1}{10} \left(9-\sqrt{161}\right)\right)\lor x=2 \pi c_1+2 \tan ^{-1}\left(\frac{1}{10} \left(9+\sqrt{161}\right)\right)\right)$$