Solving a trigonometric equation

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SUMMARY

The discussion focuses on solving the trigonometric equation 2cos(x)^2 + 3cos(x) + 1 = 0 within the interval 0 <= x <= 2π. The initial approach involved manipulating the equation to express cos(x) in terms of itself, leading to the discovery of one solution at x = π. Ultimately, the user realized that factoring the quadratic equation in cos(x) provided the necessary multiple solutions, confirming the importance of recognizing the equation's structure.

PREREQUISITES
  • Understanding of trigonometric functions and identities
  • Familiarity with quadratic equations and their solutions
  • Knowledge of the unit circle and angle measures in radians
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the factoring techniques for quadratic equations in trigonometric contexts
  • Learn about the unit circle and how it relates to trigonometric equations
  • Explore additional trigonometric identities that can simplify complex equations
  • Practice solving various trigonometric equations to enhance problem-solving skills
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Students studying trigonometry, educators teaching trigonometric equations, and anyone looking to improve their skills in solving quadratic equations involving trigonometric functions.

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Homework Statement


Solve 2cos(x)^2 + 3cos(x) + 1 = 0 for 0 <= x <= 2pi

Homework Equations


Trigonometric equations, yadda yadda yadda.

The Attempt at a Solution


2cos(x)^2 + 3cos(x) + 1 = 0
cos(x)(2cos(x) + 3) = -1
cos(x) = -1/(2cos(x) + 3)

I then figured out that you get a solution when cos(x) = -1 (so x = pi). However, there are supposed to be multiple solutions, and I don't know how to find these. Any pointers?
 
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Notice that the original equation is a quadratic in cos(x).
 
SteamKing said:
Notice that the original equation is a quadratic in cos(x).

Ah, found it. I've been messing around with trigonometric identities for half an hour now. Turns out I just had to factor the whole equation. Thanks. =D
 

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