Solve system of 5 trigonometric equations

Question

What is the best way to solve the system of following trigonometric equations with respect to $w_l, w_r, b_l, b_r$?

$$ F_x=w_l^2\sin(b_l)+w_r^2\sin(b_r) $$ $$ F_z=w_l^2\cos(b_l)+w_r^2\cos(b_r) $$ $$ M_x=w_r^2\cos(b_r)-w_l^2\cos(b_l) $$ $$ M_y=w_l^2\sin(b_l)+w_r^2\sin(b_r) $$ $$ M_z=w_r^2\sin(b_r)-w_l^2\sin(b_l) $$ I need to find relation between $w_l, w_r, b_l, b_r$ and symbolic coefficients $F_x,F_z,M_x,M_y,M_z$.

Answer 1

Note that $$F_z + M_x = 2w_r^2\cos b_r$$ $$F_x + M_z = 2w_r^2\sin b_r$$

thus for $b_r$ and $w_r$ we have (dividing one equation by another, using $\sin ^2 \theta + \cos^2 \theta = 1$):

$$\tan b_r = \frac{F_x + M_z}{F_z + M_x}$$

$$4w_r^4 = (F_x + M_z)^2 + (F_z + M_x)^2$$

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Note also that

$$F_z - M_x = 2w_l^2\cos b_l$$ $$F_x - M_z = 2w_1^2\sin b_l$$

thus for $b_l$ and $w_l$ we have

$$\tan b_l = \frac{F_x - M_z}{F_z - M_x}$$ $$ 4w_l^4 = (F_x - M_z)^2 + (F_z - M_x)^2$$

(As mentioned in the comments by OP, $M_y$ turns out to be unused indeed.)

Answer 2

If you ignore the fourth equation, since $F_x$ and $ M_y$ are identical, then number the remaining equations 1-4:

Take successively the sum of equations 1 and 4,and the difference of 1 and 4.

Call these equations A and B. Do the same operations on equations 2 and 3, call these C and D.

Now dividing B and D gives $$\tan \beta_l=\frac{F_x-M_z}{F_z-M_x}$$ Squaring and adding the same gives $$4\omega_l^4=(F_x-M_z)^2+(F_z-M_x)^2$$

Now do the same set of operations on equations A and C and you're done.