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29 September, 22:26

Which shows all the exact solutions of sin^2 x + 3cos x - 1 = 2

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  1. 29 September, 22:34
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    Answer: x = π/2 + 2πn, for n ∈ N

    Explanation:

    1) Given: sin²x + 3 cosx - 1 = 2

    2) Make sin²x = 1 - cos²x ⇒ 1 - cos²x + 3cosx - 1 = 2

    3) Add like terms and transpose terms to equal 0:

    cos²x - 3cosx + 2 = 0

    4) factor: (cos x - 2) (cosx - 1) = 0

    5) Find the solution for each factor:

    cosx - 2 = 0⇒ cosx = 2 ↔ impossible as cosx is between - 1 and 1.

    cos x - 1 = 0 ⇒ cosx = 1 ⇒ x = π/2 + 2πn, for n ∈ N
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