<s>[INST] How to solve equation with multiple trigonometric functions? Solve for $x$:
$\arccos( \cos(x) y + z) = \arcsin( \sin(x) a+b)$.
 [/INST] $$\cos^{-1}(\cos(x)y+z)=\sin^{-1}(\sin(x)a+b)$$
Taking Sine of both the sides,
$$\sin(\cos^{-1}(\cos(x)y+z))=\sin(\sin^{-1}(\sin(x)a+b))$$
$$\sin\left(\sin^{-1}\sqrt{1-(\cos(x)y+z)^2}\right)=\sin(x)a+b$$
$$\sqrt{1-(\cos(x)y+z)^2}=a\sqrt{1-\cos^2(x)}+b$$
$$1-(\cos(x)y+z)^2=\left(a\sqrt{1-\cos^2(x)}+b\right)^2$$
$$\left(1-(\cos(x)y+z)^2+a^2\cos^2(x)-a^2-b^2\right)^2=\left(2ab\sqrt{1-\cos^2(x)}\right)^2$$
$$\left((a^2-y^2)\cos^2(x)-2yz\cos(x)+1-z^2-a^2-b^2\right)^2=4a^2b^2-4a^2b^2\cos^2(x)$$
Thus, you get a quartic equation in terms of $\cos(x)$. Solve it to get value of $\cos x$
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