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Thread: [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

    $\displaystyle \cos x = \cot x$

    What I've done so far is

    $\displaystyle \cos x = \cot x$

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

    $\displaystyle \sin x = \cos x / \cos x )$

    $\displaystyle \sin x = 1$

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

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

    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 $\displaystyle \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

    $\displaystyle \cos x = \cot x$

    What I've done so far is

    $\displaystyle \cos x = \cot x$

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

    $\displaystyle \sin x = \cos x / \cos x )$

    $\displaystyle \sin x = 1$

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

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

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

    Thank you!
    $\displaystyle \cos x = \cot x$

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

    $\displaystyle \Rightarrow \sin x \cos x = \cos x$

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

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

    Case 1: $\displaystyle \cos x = 0$.

    Case 2: $\displaystyle \sin x - 1 = 0$.

    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

    $\displaystyle \cos x = \cot x$

    What I've done so far is

    $\displaystyle \cos x = \cot x$

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

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

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

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

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

    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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