Solving a trigonometric equation

AI Thread Summary
To solve the equation 2cos(x)^2 + 3cos(x) + 1 = 0 for 0 <= x <= 2pi, it is identified as a quadratic in terms of cos(x). The initial attempt involved manipulating the equation but led to confusion regarding multiple solutions. The key to finding all solutions was recognizing the need to factor the equation. Ultimately, factoring revealed the complete set of solutions, including x = pi. This approach highlights the importance of factoring in solving trigonometric equations.
KiwiKid
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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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