# How to solve this system of non linear trigonometric equations.

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?

• Note that $\omega$ in equations $3$ and $4$ is of no use. That is, if it is nonzero. – AryanSonwatikar Dec 24 '19 at 3:32

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