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

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SUMMARY

The discussion focuses on solving the equation 2bcosθ = 2sinθ - 1, where b is a constant. Participants clarify that by squaring both sides and substituting cos²θ with 1 - sin²θ, one can derive a quadratic equation in terms of sinθ. The conclusion emphasizes that transforming the original equation leads to a solvable quadratic form, confirming the necessity of this approach for finding solutions.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cos²θ = 1 - sin²θ
  • Familiarity with quadratic equations and their solutions
  • Basic algebraic manipulation skills, including squaring equations and factoring
  • Knowledge of the sine and cosine functions and their properties
NEXT STEPS
  • Study the derivation of quadratic equations from trigonometric identities
  • Practice solving quadratic equations in terms of sine and cosine
  • Explore the implications of the quadratic formula in trigonometric contexts
  • Investigate other trigonometric equations that can be transformed into quadratic forms
USEFUL FOR

Students studying trigonometry, mathematics educators, and anyone looking to deepen their understanding of solving trigonometric equations involving quadratic forms.

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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!
 

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