Question

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

Answer 1

What one is trying to do here is trying to solving a trigonometric equation `asinx+bcosx=c`. For detais please see below.

Explanation:

What one is trying to do here is trying to solving a trigonometric equation `asinx+bcosx=c`.

Dividing each term by `sqrt(a^2+b^2)`, we get the given equation

`a/sqrt(a^2+b^2)sinx+b/sqrt(a^2+b^2)cosx=c/sqrt(a^2+b^2)`

Now for solving such equation, assuming `cosalpha=b/sqrt(a^2+b^2)` and `sinalpha=a/sqrt(a^2+b^2)`.

Observe that it is compatible as `cos^2alpha+sin^2alpha=1` and `tanalpha=a/b` or `alpha=tan^(-1)(a/b)`

and then the given equation becomes

`cosxcosalpha+sinxsinalpha=c/sqrt(a^2+b^2)`

or `cos(x-alpha)=c/sqrt(a^2+b^2)`

and hence `x-alpha=2npi+-cos^(-1)(c/sqrt(a^2+b^2))`

and `x=2npi+-cos^(-1)(c/sqrt(a^2+b^2))+alpha`

or `x=2npi+-cos^(-1)(c/sqrt(a^2+b^2))+tan^(-1)(a/b)`