Solving a trigonometry equation

Question

Can anybody help me with this problem? I want to solve it for $x$ in

$$a\sin(x+θ)+b\cos(2x)=c,$$

where $a,b,c,θ$ are constants.

Answer 1

Let $\tan\frac{x}{2}=t$.

Thus, we need to solve $$a\cos\theta\sin{x}+a\sin\theta\cos{x}+b\cos^2x-b\sin^2x=c$$ or $$a\cos\theta\cdot\frac{2t}{1+t^2}+a\sin\theta\cdot\frac{1-t^2}{1+t^2}+\frac{b(1-t^2)^2}{(1+t^2)^2}-\frac{4bt^2}{(1+t^2)^2}=c$$

Answer 2

By elementary trigonometric transformations, you can rewrite the expression under the form

$$\sin^2x+a\cos x+b\sin x+c=0$$

(where my $a,b,c$ aren't yours).

Now, with

$$\sin x=\frac{2t}{1+t^2},\cos x=\frac{1-t^2}{1+t^2}$$

you obtain a general quartic polynomial equation

$$4t^2+(a+2bt-at^2)(1+t^2)+c(1+t^2)^2=0.$$

There won't be any easy shortcut.