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24 February, 17:45

Solve 2cos (x) - 4sin (x) = 3 [0,360]

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  1. 24 February, 18:09
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    2cos (x) - 4sin (x) = 3

    use identity [cos (x) ]^2 + [ sin (x) ]^2 = 1 = > cos (x) = √[1 - (sin (x)) ^2]

    2√[1 - (sin (x)) ^2] - 4 sin (x) = 3

    2√[1 - (sin (x)) ^2] = 3 + 4 sin (x)

    square both sides

    4[1 - (sin (x)) ^2] = 9 + 24 sin (x) + 16 (sin (x)) ^2

    expand, reagrup and add like terms

    4 - 4[sin (x) ]^2 = 9 + 24sin (x) + 16sin^2 (x)

    20[sin (x) ]^2 + 24sin (x) + 5 = 0

    use quadratic formula and you get sin (x) = - 0.93166 and sin (x) = - 0.26834

    Now use the inverse functions to find x:

    arcsin (-0.93166) = 76.33 degrees

    arcsin (-0.26834) = 17.30 degrees
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