Solving for Theta in 2bcosθ=2sinθ-1 Quadratic Equation

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Homework Statement


2bcosθ=2sinθ-1 where b is a constant

Homework Equations

The Attempt at a Solution


I squared both sides, foiled, and changed the cosθ2 to a 1-sin2θ. Was told I need to solve a quadratic for sine? I don't see it.
 
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Vitani11 said:

Homework Statement


2bcosθ=2sinθ-1 where b is a constant

Homework Equations

The Attempt at a Solution


I squared both sides, foiled, and changed the cosθ2 to a 1-sin2θ. Was told I need to solve a quadratic for sine? I don't see it.

How can you not end up with a quadratic in ##\sin \theta##?
 
Awesome, got it. Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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