How Do You Isolate Theta in This Trigonometric Equation?

[Meadman23]

Homework Statement

Solve for Θ

v²/RgsinΘ = sinΘ/cosΘ

Homework Equations

The Attempt at a Solution

I did:

(1) v²/RgsinΘ * (sinΘ) = sinΘ/cosΘ * (sinΘ)

which gives me:

(2) v²/Rg = sin²Θ/cosΘ

then after converting some identities I get:

(3) v²/Rg = tanΘsinΘ

I figure I can't take a tan^-1 and then a sin^-1 of the values on the left so I don't know where to go from there.

Next, starting from (2) I tried:

(3 alt) v²/Rg = (1-cos²Θ)/cosΘ

and I got:

(4) v²/Rg = secΘ - cosΘ

and it once again stalls me because I can't figure out how to get Θ alone.

My professor tells me there is a simple trig. identity that will help me figure this problem out, but I'm just not seeing one that doesn't leave me lost.

 

[Tide]

Try the following: In your Equation (2) replace $$\sin^2 \theta$$ with $$1-\cos^2 \theta$$ then solve for $$\cos \theta$$.

 

[Meadman23]

Did it, but nothing makes sense to me from my final solution that would let me get all Θ to one side.

(1) v²/Rg = 1-cos²Θ/cosΘ(2) v²cosΘ/Rg = 1-cos²Θ(3) v²cosΘ + cos²Θ = 1(4) cosΘ (v²/Rg + cosΘ) = 1 (5) v²/Rg + cosΘ = 1/cosΘ(6) v²/Rg = 1/cosΘ - cosΘ(7)v²/Rg = secΘ - cosΘ

(8)cosΘ = -v²/Rg + sec Θ

 

[Mark44]

Did it, but nothing makes sense to me from my final solution that would let me get all Θ to one side.

(1) v²/Rg = 1-cos²Θ/cosΘ

The above should read v²/Rg = (1-cos²Θ)/cosΘ

I.e., you need parentheses.

(2) v²cosΘ/Rg = 1-cos²Θ


(3) v²cosΘ + cos²Θ = 1

The next step is to move everything to one side and rearrange a bit. I also put Rg back in, since you seemed to have lost it.
cos²Θ + v²cosΘ/(Rg) - 1 = 0

The equation is now quadratic in form, and can be solved for cos(Θ). From there you can solve for Θ.

(4) cosΘ (v²/Rg + cosΘ) = 1


(5) v²/Rg + cosΘ = 1/cosΘ


(6) v²/Rg = 1/cosΘ - cosΘ


(7)v²/Rg = secΘ - cosΘ

(8)cosΘ = -v²/Rg + sec Θ