How to simplify the equation $ a\sin x\cos y + b\cos x\sin y = c\sin(2x)$?

Question

I am trying to derive the relation between x and y which are two physical parameters.I am trying to express x in terms of y, a and b. Is there a way to do it?

Answer 1

The equation is of the form

$$p(x)\cos y+q(x)\sin y=r(x),$$ which is a classical linear trigonometric equation.

Let $t:=\cos y$. Then rewrite

$$q^2(1-t^2)=(r-pt)^2$$

which is a quadratic equation in $t$.

Finally $$\cos y=\frac{a\sin x\,c\sin 2x\pm b\cos x\sqrt{a^2\sin^2x+b^2\cos^2x-\,c^2\sin^22x}}{a^2\sin^2x+b^2\cos^2x}.$$

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As shown by @Blue, the converse relation leads to a quartic problem, a little harder, though it has a (slightly monstrous) closed-form solution.

See https://www.wolframalpha.com/input/?i=ax%5E4%2Bbx%5E3%2Bcx%5E2%2Bdx%2Be%3D0

Answer 2

You can rewrite your equation as $$b \cos x \sin y = \sin x \;( 2 \cos x - a \cos y ) $$ Squaring, we have $$b^2 \cos^2 x \sin^2 y = \sin^2 x (2 \cos x - a \cos y )^2$$ So, writing $p:= \cos x$ and $q := \cos y$, $$b^2 p^2 (1-q^2) = (1-p^2) (2p - a q )^2$$ This gives a quartic polynomial in $p$: $$4 c^2 p^4 - 4 a c p^3 q + ( b^2 - 4 c^2 + a^2 q^2 - b^2 q^2 ) p^2 + 4 a c p q - a^2 q^2 = 0 $$ which does not have nice roots.