Is There a Trick to Solve for Theta in this Algebraic Equation?

[scavok]

http://home.comcast.net/~andykovacs/equation.GIF

g is a constant. I need to find theta.

Is there some trick I can do to cancel out the denominator in the root?

 

[arildno]

Hint:
Set
$$y=\sqrt{\sin\theta}$$

and see what type of equation you get for y.

 

[HallsofIvy]

http://home.comcast.net/~andykovacs/equation.GIF

g is a constant. I need to find theta.

Is there some trick I can do to cancel out the denominator in the root?

No, you can't cancel- but you can square. I would first divide the entire equation by $g sin(\theta)$ to get
$$\frac{20}{g sin(\theta)}= \frac{14}{g sin(\theta)}+ \sqrt{\frac{14}{g sin(\theta)}}+ 8.41$$
I would even write it as
$$\frac{10}{7}\frac{14}{g sin(\theta)}= \frac{14}{g sin(\theta)}+ \sqrt{\frac{14}{g sin(\theta)}}+ 8.41$$
because then I can let $y= \frac{14}{g sin(\theta)}$ and have
$$\frac{10}{7}y= y+ \sqrt{y}+ 8.41$$
That gives
$$\frac{3}{7}y- 8.41= \sqrt{y}$$
Square on both sides:
[tex](\frac{9}{49}y- 8.41)^2= y[/itex]<br /> <br /> Solve that quadratic equation for y and then solve <br /> $$y= \frac{14}{g sin(\theta)}$$<br /> for $\theta$.[/tex]

 

[scavok]

Do you get y=-11.2432 and y=-1.0611 after solving the quadratic equation? If so then I probably screwed up with the physics.