Solve for Θ on the interval [0,360)?

Question

2 sin2Θ - 7 = 4cscΘ

Answer 1

I'm not sure what you are asking for.

There are several possible approaches. Assuming we are looking for a numerical solution, there are two obvious ones:

1. Convert the equation to a polynomial

2. Solve it directly

Let's start with converting the equation to a polynomial:

Using the trigonometric identities, we have sin2Θ = 2 sinΘ cosΘ

http://en.wikipedia.org/wiki/Trigonometric_identity

So adding 7 to both sides and then squaring, we get sin^2 and cos^2 on the left which allows us to substitute (1 - sin^2) for cos^2.

Then use the definition of cosecant to get: cscΘ = 1/sinΘ

http://en.wikipedia.org/wiki/Cosecant

Setting u = sinΘ, we have a 4th degree polynomial in u. Solve by whatever method you like and compute the arcsin of the result.

Quartics have explicit formulas:
http://en.wikipedia.org/wiki/Quartic_equation
http://home.att.net/~srschmitt/script_quartic.html

But you may prefer a more general approach:
http://www.crbond.com/roots.htm

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For solving the problem directly, I'd use Newton's method to solve the equation in radians and then convert:
http://en.wikipedia.org/wiki/Newton's_method

The important element in using Newton's method is starting with a good guess.

Since the sine has a maximum value of 1, the left hand side is going to be between -5 and -9.

That means that the cosecant has to have a value of between -5/4 and -9/4. If we choose -8/4 as our initial guess, this gives us a Θ of -30 degrees or -120 degrees (-pi/6 or -7pi/6).

For Θ = -30, sin2Θ is negative, while for -120 it is positive. That means that -30 is closer so I'd use that (i.e. 330 degrees) as my initial guess.

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If you've already gotten this far and have other problems, or if you need something completely different, I suggest you post a clarification.