Factoring Trigonometric Expressions...3

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Dec 10, 2021.

  1. nycmathguy

    nycmathguy

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    Section 5.1

    Screenshot_20211207-195319_Samsung Notes.jpg

    Question 30

    6 cos^2 x + 5 cos x - 6

    Let u = cos x

    6u^2 + 5u - 6

    Factor by grouping.

    6(-6) = -36

    9(-4) = -36

    9 + (-4) = 5

    6u^2 + 9u - 4u - 6

    6u^2 + 9u becomes 3u(2u + 3)

    -4u - 6 becomes -2(2u + 3)

    We now have (3u - 2)(2u + 3).

    Back-substitute for u.

    (3cos x - 2)(2cos x + 3)

    You say?
     
    nycmathguy, Dec 10, 2021
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  2. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Dec 10, 2021
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  3. nycmathguy

    nycmathguy

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    Thanks.
     
    nycmathguy, Dec 10, 2021
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