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Math Help - simple trig equation

  1. #1
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    simple trig equation

    ok so i have been doing these trig equations for a while now...and i came up with this suspicious easy one..can anybody tell me if what im doing is right?

    cos^2x = 1

    took the square root of both sides and ended up with cos x = 1... then i took the inverse of cos (1) and came up with 0.. now im checking when cos is positive, and it turns out its in Q 1 and 4

    so in Q1 its the same as the related acute which is 0 and in Q 4 its 2 pi - 0 which gives 2 pi...

    is this right? is it ok to take the square root of both sides to remove the squared in the cosine??
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by magmagod View Post
    ok so i have been doing these trig equations for a while now...and i came up with this suspicious easy one..can anybody tell me if what im doing is right?

    cos^2x = 1

    took the square root of both sides and ended up with cos x = 1... then i took the inverse of cos (1) and came up with 0.. now im checking when cos is positive, and it turns out its in Q 1 and 4

    so in Q1 its the same as the related acute which is 0 and in Q 4 its 2 pi - 0 which gives 2 pi...

    is this right? is it ok to take the square root of both sides to remove the squared in the cosine??
    You missed some stuff...

    if \cos^2(x) = 1, \cos(x) = \pm 1.

    Your answers of 0 and 2 \pi were correct, however, it will also be equal to 1 at \pm 2\pi, \pm 4\pi, etc...

    So \cos(x) = 1 when x = \pm 2n\pi, n = 0,1,2,3, \dots.

    But we also have solutions when \cos(x) = -1, this happens at \pm \pi, \pm \frac{3\pi}{2}, \pm \frac{5\pi}{2}, etc...

    I.e. when x = \pm \frac{(2n+1)\pi}{2}, n=0,1,2,3,\dots.
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  3. #3
    Super Member Deadstar's Avatar
    Joined
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    Here's a graph of \cos^2(x). Not.e where the graph is equal to 1. i.e. at the top of the 'waves'.
    Attached Thumbnails Attached Thumbnails simple trig equation-untitled.jpg  
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