# Homework Help: Solve the equation and find x

1. Nov 14, 2015

### skrat

1. The problem statement, all variables and given/known data
I ran across this equation: $$T_1\cos \vartheta =-T_2\cos \alpha \sin \varphi + T_2\sin \alpha \cos \varphi$$
Get $\alpha$ if $T_1=80$, $T_2=50$, $\vartheta =60^°$ and $\varphi =30^°$.

2. Relevant equations

3. The attempt at a solution
There are two possible ways:
a) One could simply write $$\sin^2=1-\cos^2\alpha$$ and get a quadratic equation for $\cos \alpha$ with solutions $$(\cos\alpha)_1=0.1196$$ and $$(\cos\alpha)_2=-0.9196.$$ Each solution has two angles $\alpha$. First solution is $$\alpha_1=\pm 83.13^°$$ and second $$\alpha _2=\pm 156.87^°$$

b) One could use some trigonometric identities and rewrite the equation as $$T_1 \cos\vartheta = T_2\sin(\alpha-\varphi)$$ which means (according to my understanding...) $$\alpha =\varphi+arcsin(\frac{T_1}{T_2}\cos\vartheta)+n\pi.$$ For $n=0$ the solution is (as above in case a)) $\alpha_1= 83.13^°$, however the problem here is that I can't figure out how to get the other solution in this case.

This is absolutely the most embarrassing question when you realize you are close to a second Bachelor degree...

2. Nov 14, 2015

### Samy_A

$\sin(\pi - x)=\sin(x)$
That should give you the other solution.
Remember that one has to restrict the range of the inverse trigonometric functions in order to have them as well defined functions.

3. Nov 14, 2015

### vela

Staff Emeritus
The $n\pi$ part is wrong. Consider, for example, if $\varphi = 0$ and $\frac{T_1}{T_2}\cos\vartheta = 1$. Then you'd have $\sin \alpha = 1$. One solution is $\alpha = \frac{\pi}{2}$, but clearly $\alpha = -\pi/2$ (n = -1) and $\alpha = 3\pi/2$ (n=1) aren't solutions.

4. Nov 14, 2015

### Ray Vickson

If you have $0 < v < 1$ and an equation $\sin(\theta) = v$, the two smallest positive roots are $\theta_1 = \arcsin(v)$ and $\theta_2 = \pi - \theta_1$.

Just look at the graph of $y = \sin(\theta)$ for $0 \leq \theta \leq \pi$. You will see that the horizontal line $y = v$ cuts the sine-graph at two points, $\theta_1$ and $\pi - \theta_1$. Alternatively, look at the unit circle in the $(x,y)$ plane, and notice that points at angles $\theta$ and $\pi - \theta$ have the same y-coordinate.