# Solve a system of trigonometric equations

How can I solve this system of trigonometric equations analytically? It is from physics class. $$\begin{cases} 30t\cos{\alpha}=50\\ -30t\sin{\alpha}-4.9t^2=0 \end{cases}$$

• Hint: try moving the $t^2$ term to the other side, squaring the two equations, and adding them together. What happens to the trigonometric terms? – Josh B. Nov 28 '18 at 17:16

Hint: Squaring both the equations, you will get $$900t^2\cos^2{\alpha}=2500\\ 900t^2\sin^2{\alpha}={4.9}^2t^4$$.

Note that $$\sin^2{\alpha}+\cos^2{\alpha}=1$$.

So add both the equations and solve for $$t$$ using the substitution $$t^2=u$$.

• Turn it into a quadratic is the gist of my answer but this is much nicer+faster. – Mason Nov 28 '18 at 17:22
• @Mason Thank you. – Thomas Shelby Nov 28 '18 at 17:24

$$30t\cos{\alpha}=50 \implies t=\frac{5}{3} \sec\alpha$$

You can plug this information into the other equation and solve:

$$-30t\sin{\alpha}-4.9t^2=0\implies -30(\frac{5}{3} \sec\alpha)\sin{\alpha}-4.9(\frac{5}{3} \sec\alpha)^2=0$$

$$-50(\tan\alpha)-4.9(\frac{5}{3} \sec\alpha)^2=0$$

$$-50(\tan\alpha)-4.9\frac{25}{9} \sec^2\alpha=0$$

$$-50(\tan\alpha)-4.9\frac{25}{9} (\tan^2\alpha+1)=0$$

Taking $$y=\tan\alpha$$ you can solve a quadratic equation.
$$-50(y)-4.9\frac{25}{9} (y^2+1)=0$$

I think you're probably in good shape from here?

• This results in negative alpha but its a real physics problem, so it must be positive. – Stepii Nov 28 '18 at 17:27
• The angle cannot be negative? Add $2\pi$? – Mason Nov 28 '18 at 17:29
• Adding 2π gives angles greater than 90° – Stepii Nov 28 '18 at 17:38
• @Stepii Which also doesn't make sense in the context of the problem? – Mason Nov 28 '18 at 17:43
• What did you get for $y$. Maybe I made an error... Here's y – Mason Nov 28 '18 at 17:44

Sorry. This is not an answer. Just part of a chat conversation that we are not really suppose to be having via comments.

This is the image from the link you shared. Unless I am reading this graph wrong it says: That when $$a<0$$ then $$t>0$$

• The extraneous solutions are when $a>0$ because it implies that $t<0$ which as you have commented makes little sense often in physics problems. – Mason Nov 28 '18 at 18:23
• But that solutions (with the angle and time both positive) satisfy the given system – Stepii Nov 28 '18 at 18:29