Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

I have a single trig problem I do not understand, could you please explain how I can solve this?  ?

Here's the question

Solve algebraically cos(θ)+2cos(θ)+1=0 with exact solutions in the interval [0, 2π)

A walkthrough would be really apricated

2 Answers

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  • 1 month ago
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    Really?  Are you sure you didn't mean cos^2(theta) + 2*cos(theta) + 1 = 0 ?

    The equation you actually wrote means 3cos(theta) = -1, so 

    theta = arccos(-1/3), one solution in the 2nd quadrant and one in the 3rd quadrant, but neither of them any rational number of degrees.

    If you meant cos^2(theta) + 2*cos(theta) + 1 = 0, then it's the same as

    [cos(theta) + 1] ^ 2 = 0, so

    cos(theta) = -1, which implies theta = pi.

  • 1 month ago

    cos(θ) + 2*cos(θ) + 1 = 0

    3*cos(θ) + 1 = 0

    3*cos(θ) = -1

    cos(θ) = -1 /3

    General solution (exact):

    θ = 2*pi*n +/- arccos(-1/3), for any integer n

    Approximate: 

    https://www.wolframalpha.com/input/?i=cos%28%CE%B8...

    Exact solutions in the interval 0 <= θ <= 2*pi:

    θ = arccos(-1/3) or θ = 2*pi - arccos(-1/3)

    https://www.wolframalpha.com/input/?i=cos%28%CE%B8...

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