Find $x$ in terms of the constants: $a$, $b$, $c$, and $d$. $$\sin(\arcsin(a-x) -c) = \sin(\arcsin(b+x)+\pi-c) + d$$
(Assume $d$ is around $.01$-$.21$.)
I don't how to solve this problem or if it is even possible to solve. Please, help me solve it or show me why it is an impossible math question.
Thanks
Use the identity $\sin(x+y) = \sin x\cos y+\cos x\sin y$ to expand the equation,
$$\sin[\arcsin(a-x) -c] = \sin[\arcsin(b+x)+\pi-c] + d$$
to
$$(a-x)\cos c-\sqrt{1-(a-x)^2}\sin c = -(b+x)\cos c+\sqrt{1-(b+x)^2}\sin c + d$$
where $\cos (\arcsin t) = \sqrt{1-t^2}$ is used. Rearrange the resulting equation to obtain the following,
$$\sqrt{1-(a-x)^2} = r-\sqrt{1-(b+x)^2}$$
where
$$r = \frac{(a+b)\cos c-d}{\sin c}$$
Then, square both sides (twice) to remove square-roots and you end up with a standard quadratic equation in $x$.