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2sin^2(x )+38 cos(x) + 2.6=0

  • Trigonometry -

    Change Sin^2(x) into 1-cos^2(x), substitute it in, tidy it up and you'll get a quadratic of the form
    aCos^2(x) + bCos(x) + c = 0

    Then let, say, y = Cos(x) then and you get

    ay^2 + by + c = 0
    which is a more normal quadratic. Solve that for y and put your answer equal to Cos(x). There will be 2 possible values of y probably and 2 possible values for Cos (x) for each value of y. In total, 4 answers.

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