Solve Trig Inequality: 2cos^2(x) + 1 = 3cos(2x) [0,2pi)

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SUMMARY

The discussion focuses on solving the trigonometric inequality 2cos²(x) + 1 = 3cos(2x) within the interval [0, 2π]. Participants emphasize the importance of using the identity cos(2x) = 2cos²(x) - 1 to rewrite the equation correctly. The initial attempts included incorrect manipulations of the cosine terms, leading to confusion. The correct approach simplifies the equation to a solvable polynomial form.

PREREQUISITES
  • Understanding of trigonometric identities, specifically cos(2x) = 2cos²(x) - 1
  • Knowledge of polynomial equations and their solutions
  • Familiarity with the unit circle and the interval [0, 2π]
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation and applications of trigonometric identities
  • Learn how to solve polynomial equations derived from trigonometric functions
  • Practice solving inequalities involving trigonometric functions
  • Explore graphical methods for visualizing trigonometric equations
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Students studying trigonometry, educators teaching trigonometric equations, and anyone looking to enhance their problem-solving skills in trigonometric inequalities.

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Homework Statement


solve the following equations or inequalties for x in the interval [0,2pi)

2cos^2(x) + 1 = 3cos(2x)

Homework Equations





The Attempt at a Solution



My attempt at the problem:

2cosx(cos2x) + 1 = 3cos(2x)
2cosx(cos^2x - sin^2x) + 1 = 3cos(2x)
2cos^3x - 2cosxsin^2x + 1 = 3(1-2sin^2x)
2cos^3x - 2cosxsin^2x + 1 = 3-6sin^2x
2cos^3x - 2cosxsin^2x + 1 - 3 + 6sin^2x = 0

and I am getting lost here...


please help,
 
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use this identity:
Cos(2x) = 2Cos^2(x) - 1
 
lovemake1 said:

Homework Statement


solve the following equations or inequalties for x in the interval [0,2pi)

2cos^2(x) + 1 = 3cos(2x)


My attempt at the problem:
You have a mistake in your first step, below. 2cos2(x) is not equal to 2cos(x)*cos(2x). Use the identity that JonF gave to rewrite 3cos(2x) in terms of cos2(x).
lovemake1 said:
2cosx(cos2x) + 1 = 3cos(2x)
2cosx(cos^2x - sin^2x) + 1 = 3cos(2x)
2cos^3x - 2cosxsin^2x + 1 = 3(1-2sin^2x)
2cos^3x - 2cosxsin^2x + 1 = 3-6sin^2x
2cos^3x - 2cosxsin^2x + 1 - 3 + 6sin^2x = 0

and I am getting lost here...


please help,
 

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