How to solve this system of non linear trigonometric equations.

Question

How to solve this system of non linear trigonometric equations:

$$\begin{align} A\sin\theta_1+\phantom{5\omega}B\sin\theta_2 &=P \tag{1}\\ 2A\sin\theta_1+\phantom{\omega}5B\sin\theta_2 &=Q \tag{2}\\ A\omega\cos\theta_1+\phantom{5}B\omega\cos\theta_2 &=0 \tag{3}\\ 2A\omega\cos\theta_1+5B\omega\cos\theta_2 &=0 \tag{4} \end{align}$$

$A$, $B$, $\theta_1$, $\theta_2$ are variables, and $\omega$ is a constant.

Can you at least give me a hint on how to proceed?

Answer 1

Following assumptions are to be made: $A, B,$ and $\omega \ne 0$
$$\left(2\right)-2\times\left(1\right)\space\space\space\space 3B\sin\theta_2=Q-2P$$ $$\left(4\right)-2\times\left(3\right)\space\space\space\space\space 3B\cos\theta_2=0\space\space\space\space\space\space\space\space\space\space\space$$ $$5\times\left(1\right)-\left(2\right)\space\space\space\space 3A\sin\theta_1=5P-Q$$ $$5\times\left(3\right)-\left(4\right)\space\space\space\space\space 3A\cos\theta_1=0\space\space\space\space\space\space\space\space\space\space\space$$ This is the hint you requested. Now please proceed.

Answer 2

Hint:

Use $(3)$ to simplify $(4)$ and $(1)$ to simplify $(2)$. Square and add to eliminate $\theta_2$. Can you proceed?