# 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 $$\sin \alpha =-4/5$$?

• I would say $$-2\sin x + \cos x =1$$ is equivalent to $$\sqrt 5\cdot \cos \left( x + \arctan 2\right) =1$$ or to $$\sqrt 5\cdot \sin \left( x - \arctan \frac 12\right) =-1$$ – dfnu Mar 2 at 13:21
• It could help to realize that $\arctan\frac12 = \arcsin\frac{\sqrt{5}}{5}$. – Blue Mar 2 at 13:22
• @dfnu In the meantime, thank you. I did not understand the reason in RHS because you wrote $-1$ when in RHS I have $1$. Everything else is clear. – Sebastiano Mar 2 at 21:44