Linear trigonometric equation

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

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

or

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

Now, if

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

we have a countable set of solutions

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

or

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

for $ k\in Z $

If

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

there are no real solutions.

Average: 4 (6 votes)

Back to top