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Precalculus 2

posted by on .

Similiarly to a question I asked previously I took the same approach. the problem is:

what value(s) of theta solve the following equation? cos(theta)sin(theta) - 2 cos(theta) = 0?

I let Cos theta = X and sin theta = X
unfortunately I end up with something completely odd. Can anyone help me out please?

  • Precalculus 2 - ,

    cosØ sinØ - 2cosØ = 0
    factor it
    cosØ(sinØ - 2) = 0
    cosØ = 0 ----> Ø = π/2 , 3π/2 (90°, 270°)
    or
    sinØ = 2 , which has no solution

    so

    Ø = π/2 or 3π/2

    BTW, how can you let both cosØ and sinØ = x ???
    that would be true only if cosØ = sinØ

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