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Math Help - [SOLVED] Solving Trigonometric Functions algebraically

  1. #1
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    [SOLVED] Solving Trigonometric Functions algebraically

    Hello, first time posting.

    Had a quick question on a problem I was doing.

    The problem:

    Find exact solutions

    \cos x = \cot x

    What I've done so far is

     \cos x = \cot x

    \cos x = (\cos x / \sin x)

    \sin x = \cos x / \cos x )

    \sin x = 1

    Now, when \sin x = 1 then x is \pi / 2. But the answer is also  3 \pi / 2

    But isn't 3 \pi / 2 = -1?

    I'm missing a part of the theory obviously, was just wondering if someone could clarify.

    Thank you!
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  2. #2
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    but  \cos{\frac{3\pi}{2}} = 0
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  3. #3
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    When cosx is positive, cotx may be positive or negative.
    So square both the sides, you get
    cos^x = cot^2x
    Cos^2x = 0 or sin^2x = 1 or sinx = (+ or -) 1
    Hence x = 90 οr 270 degrees
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  4. #4
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    Quote Originally Posted by sleels View Post
    Hello, first time posting.

    Had a quick question on a problem I was doing.

    The problem:

    Find exact solutions

    \cos x = \cot x

    What I've done so far is

     \cos x = \cot x

    \cos x = (\cos x / \sin x)

    \sin x = \cos x / \cos x )

    \sin x = 1

    Now, when \sin x = 1 then x is \pi / 2. But the answer is also  3 \pi / 2

    But isn't 3 \pi / 2 = -1?

    I'm missing a part of the theory obviously, was just wondering if someone could clarify.

    Thank you!
    \cos x = \cot x

    \Rightarrow \cos x = \frac{\cos x}{\sin x}

    \Rightarrow \sin x \cos x = \cos x

    \Rightarrow \sin x \cos x - \cos x = 0

    \Rightarrow \cos x (\sin x - 1) = 0.

    Case 1: \cos x = 0.

    Case 2: \sin x - 1 = 0.

    Solve each case over the given domain.

    By cancelling the cos x in your solution you fell into the classic trap for young players.
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  5. #5
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    Quote Originally Posted by sleels View Post
    Hello, first time posting.

    Had a quick question on a problem I was doing.

    The problem:

    Find exact solutions

    \cos x = \cot x

    What I've done so far is

     \cos x = \cot x

    \cos x = (\cos x / \sin x)

    \sin x = \cos x / \cos x )
    This division is valid only if cos x is not equal to 0!

    \sin x = 1
    And if sin x= 1 (or -1), then cos x= 0! Your whole method of solution is invalid.

    Now, when \sin x = 1 then x is \pi / 2. But the answer is also  3 \pi / 2

    But isn't 3 \pi / 2 = -1?

    I'm missing a part of the theory obviously, was just wondering if someone could clarify.

    Thank you!
    Far better to use mr. fantastic's method, which does not depend on dividing by a function.
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  6. #6
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    Thank you everyone for the help. I see what I was doing wrong now!
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