# How do we solve the equation asinx+bcosx=c?

Oct 17, 2017

What one is trying to do here is trying to solving a trigonometric equation $a \sin x + b \cos x = c$. For detais please see below.

#### Explanation:

What one is trying to do here is trying to solving a trigonometric equation $a \sin x + b \cos x = c$.

Dividing each term by $\sqrt{{a}^{2} + {b}^{2}}$, we get the given equation

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

Now for solving such equation, assuming $\cos \alpha = \frac{b}{\sqrt{{a}^{2} + {b}^{2}}}$ and $\sin \alpha = \frac{a}{\sqrt{{a}^{2} + {b}^{2}}}$.

Observe that it is compatible as ${\cos}^{2} \alpha + {\sin}^{2} \alpha = 1$ and $\tan \alpha = \frac{a}{b}$ or $\alpha = {\tan}^{- 1} \left(\frac{a}{b}\right)$

and then the given equation becomes

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

or $\cos \left(x - \alpha\right) = \frac{c}{\sqrt{{a}^{2} + {b}^{2}}}$

and hence $x - \alpha = 2 n \pi \pm {\cos}^{- 1} \left(\frac{c}{\sqrt{{a}^{2} + {b}^{2}}}\right)$

and $x = 2 n \pi \pm {\cos}^{- 1} \left(\frac{c}{\sqrt{{a}^{2} + {b}^{2}}}\right) + \alpha$

or $x = 2 n \pi \pm {\cos}^{- 1} \left(\frac{c}{\sqrt{{a}^{2} + {b}^{2}}}\right) + {\tan}^{- 1} \left(\frac{a}{b}\right)$