Trig Equation: Solving for Theta with Tangent and Sine Functions

[lylos]

Homework Statement

Tan[60 degrees]=Sin[theta]/(1/3+Cos[theta])

Homework Equations

Trig identities?

The Attempt at a Solution

Tan[60 degrees]=Sin[theta]/(1/3+Cos[theta])
(1/3+Cos[theta])Tan[60 degrees]=Sin[theta]
1/3 Tan[60 degrees]=Sin[theta]-Tan[60 degrees]*Sin[theta]

Now I'm lost...
I can put it in mathematica, and get a solution of 1.34004 Radians or 76.7786 degrees. I just don't know how to solve it!

 

[quantumdude]

The Attempt at a Solution

Tan[60 degrees]=Sin[theta]/(1/3+Cos[theta])
(1/3+Cos[theta])Tan[60 degrees]=Sin[theta]

OK so far.

1/3 Tan[60 degrees]=Sin[theta]-Tan[60 degrees]*Sin[theta]

The line is wrong. Your cosine morphed into a sine, so you'll need to fix that. I would actually back up to the previous line, leaving the cosine and sine on different sides. Look at \sin(\theta) on the right side. Use a Pythagorean identity to express it in terms of \cos(\theta).

 

[lylos]

Thank you, I was able to get to the solution!