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

1 Answer
Oct 17, 2017

Answer:

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)#