Question

Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 ≤ x ≤ 360°. Use exact solutions whenever possible. 6cos2(θ) + cos(θ) - 1 = 0 2 - 2sin2(θ) = cos(θ)Solve the following quadratic trig equations algebraically (without graphing technology) for the domain 0 ≤ x ≤ 360°. Use exact solutions whenever possible.
A. 6cos2(θ) + cos(θ) - 1 = 0
B. 2 - 2sin2(θ) = cos(θ)

Answer 1

Found the solution, no worries.
For those of you looking for it:

A. 6cos(Theta) + cos(Theta) - 1 = 0

[2cos(Theta) + 1] [3cos(Theta) - 1] = 0

2cos(Theta) + 1 = 0 or 3cos(Theta) - 1 = 0

cos(Theta) = -1/2 or cos(Theta) = 1/3

(Theta )=180-60 or (Theta)=180+60 or (Theta)=70.5 or (Theta)=360-70.5

(Theta) = 120 or (Theta) = 240 or (Theta) = 70.5 or (Theta) = 289.5


b. 2 - 2sin(Theta) = cos(Theta)

2[sin(Theta) + cos(Theta)] - 2sin(Theta) = cos(Theta)

2sin(Theta) + 2cos(Theta) - 2sin(Theta) = cos(Theta)

2cos(Theta) - cos(Theta) = 0

cos(Theta) [2cos(Theta) - 1] = 0

cos(Theta) = 0 or cos(Theta) =

(Theta) = 90 or (Theta) = 270 or (Theta) = 60 or (Theta) = 360-60

(Theta) = 90 or (Theta) = 270 or (Theta) = 60 or (Theta) = 300

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Found the solution here: http://hk.knowledge.yahoo.com/question/question?qid=7008071400443