Linear trigonometric equation

Submitted by Structure on Wed, 09/24/2008 - 06:18.

We solve the equation

$a\sin x+b\cos x+c=0$

where $a,b,c \in R$ and $a^2+b^2>0$ .
Then we can write

$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin x+\frac{b}{\sqrt{a^2+b^2}}\cos x+\frac{c}{\sqrt{a^2+b^2}})=0$

There is a unique $t\in[0,2\pi)$ such as

$\cos t=\frac{a}{\sqrt{a^2+b^2}}$

and

$\sin t=\frac{b}{\sqrt{a^2+b^2}}$

.
so we can write

$\sqrt{a^2+b^2}(\sin x\cos t+\cos x\sin t+\frac{c}{\sqrt{a^2+b^2}})=0$

$\sqrt{a^2+b^2}(\sin (x+t)+\frac{c}{\sqrt{a^2+b^2}})=0$

Now, if

$|\frac{c}{\sqrt{a^2+b^2}}|\leq 1$

we have a countable set of solutions

$x_k+t=k\pi+(-1)^{k+1}\arcsin\frac{c}{\sqrt{a^2+b^2}}$

$x_k=-t+k\pi+(-1)^{k+1}\arcsin\frac{c}{\sqrt{a^2+b^2}}$

for $k\in Z$

$|\frac{c}{\sqrt{a^2+b^2}}|> 1$

there are no real solutions.