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Math Help - Trig question: with intervals

  1. #1
    VkL
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    Trig question: with intervals

    Solve the equation for a in [0, 2pie]

    A.) cosē(a) + 3cos(a) = -cos(a)




    B.)sin(2a) = cos(a)
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  2. #2
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    Quote Originally Posted by VkL View Post
    Solve the equation for a in [0, 2pie]

    A.) cosē(a) + 3cos(a) = -cos(a)
    Add \cos a to both sides to get \cos^2 a + 4\cos a = 0 and now \cos a ( \cos a + 4) = 0. Therefore, \cos a = 0 or \cos a + 4 = 0. The second condition, \cos a + 4 = 0 \implies \cos a = -4 is impossible because |\cos a| \leq 1. The first condition is \cos a = 0. That happens when a = \tfrac{\pi}{2},\tfrac{3\pi}{2}


    B.)sin(2a) = cos(a)
    Using the double angle identity we get 2\sin a \cos a = \cos a \implies 2\sin a \cos a - \cos a = 0. Now factor, \cos a ( 2\sin a - 1) = 0. Thus, we get two possibilities, \cos a = 0 or 2\sin a - 1 = 0. The first condition, \cos a = 0, gives a = \tfrac{\pi}{2}, \tfrac{3\pi}{2}. The second condition, 2\sin a - 1 = 0 is equivalent to saying \sin a = \tfrac{1}{2}, thus, a = \tfrac{\pi}{6}, \tfrac{5\pi}{6}.
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