Solution of a trigonometric linear equation with the method of the added angle This morning I have done for my students of an high school an exercise with the method of the added angle. I not write all the steps but the principals.
The equation is:
$$\bbox[5px,border:3px solid #FF7F50]{-2\sin x+\cos x = 1} \tag 1$$
Surely $x=2\Bbb Z \pi$ it is solution of the $(1)$. In fact it is an identity. Being $$\tan \varphi=-\frac 12 \implies \varphi =-\arctan \frac 12, \quad A=\sqrt{5}$$
Now the $(1)$ becomes
$$\bbox[5px,border:3px solid #AA7F50]{\sqrt 5 \sin \left(x-\arctan \frac 12\right)=1} \tag 2$$
Hence I will have the other two solutions:
$$x=\arctan \frac 12+\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi$$ and $$ \quad x=\pi +\arctan \frac 12-\arcsin\left( \frac{\sqrt{5}}5\right)+2\Bbb Z\pi \tag 3$$
The solution of my textbook are $x=k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$.

Please, do can be that the $(3)$ are equivalent to $x=2k\pi$ and $x=\alpha+2k\pi$ where $\cos \alpha =-3/5$ and $\sin \alpha =-4/5$?


Putting together the two comments we have that the initial equation is equivalent to
$$\sqrt 5 \sin\left(x - \arctan \frac12\right) = -1$$
Leading to the solutions
$$ x = \arctan \frac12 - \arcsin \frac{\sqrt 5}5 + 2k \pi = 2k \pi$$
and
$$x = \pi + \arcsin \frac{\sqrt 5}5 + \arctan \frac12 + 2k \pi=\underbrace{\pi + 2 \arcsin \frac{\sqrt 5}5}_{\alpha}+ 2 k \pi.\tag{1} \label{1}$$
Now let $\alpha = \pi + 2\arcsin \frac{\sqrt 5}5$. We get
$$\sin \alpha = -\sin\left( 2\arcsin \frac{\sqrt 5}5\right) = -2 \sin\left( \arcsin \frac{\sqrt 5}5\right)\cos\left(\arcsin\frac{\sqrt 5}5\right)=-\frac45,$$
and similarly for the cosine of $\alpha$.

Note that $k \pi$ is not a solution for odd $k$.
