Generic way to solve $ a \cos(x) + b\sin(x) = c $

Question

Is there a generic way without using complex numbers to solve equations of the form:

$$ a \cos(x) + b\sin(x) = c $$

Answer 1

Compare your equation to $R\cos (x - \theta) = R\cos x\cos\theta + R\sin x \sin\theta$

By comparing coefficients, deduce that:

$R\cos\theta = a$

and $R \sin \theta = b$

If you take the second equation divided by the first, you get:

$\tan \theta = \frac ba \implies \theta = \arctan \frac ba$

If you square the two equations and add them up you get:

$R^2 (\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \implies R^2 = a^2 + b^2 \implies R = \sqrt{a^2 + b^2}$

You can now cast the original equation into the form:

$\sqrt{a^2 + b^2}\cos(x - \arctan \frac ba) = c$

and I hope the rest of the solution is obvious.

Answer 2

There is a general solution of the form

$$a cos\theta + b sin\theta = Rcos(\theta-\alpha)$$ $$a cos\theta - b sin\theta = Rcos(\theta+\alpha)$$

where $$\alpha=tan^{-1}\frac{b}{a}$$

$$R=(a^2+b^2)^{\frac{1}{2}}$$

Answer 3

We have to solve,

$\qquad a \cos (x) + b \sin(x)$

Dividing the equation by $ \sqrt{a^2+b^2}$,

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

Let $\cos(r)=\dfrac{a}{\sqrt{a^2+b^2}}$

Then the equation becomes,

$ \cos(x-r)=\dfrac{c}{\sqrt{a^2+b^2}}$

Let $\cos(α) = \dfrac{c}{\sqrt{a^2+b^2}}$

Then,

$x-r = 2nπ±\alpha$

And, $ \qquad x = 2nπ±\alpha +r $

That, how we can solve these type of trigonometric equations.