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Thread: tan equation

  1. #1
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    Smile tan equation

    Solve for 0<x<360, the equation tan(2x - 45) = 1

    (N.B. the question is in degrees & the interval is greater than or EQUAL to 0 and less than OR EQUAL to 360)

    The solution is:
    x = 45, 135, 225, 315

    I am just not sure how to get to the answer-help would really be appreciated- BIG exam in less than a day. Thanks x
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  2. #2
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    Quote Originally Posted by orangesmarties View Post
    Solve for 0<x<360, the equation tan(2x - 45) = 1

    (N.B. the question is in degrees & the interval is greater than or EQUAL to 0 and less than OR EQUAL to 360)

    The solution is:
    x = 45, 135, 225, 315

    I am just not sure how to get to the answer-help would really be appreciated- BIG exam in less than a day. Thanks x
    $\displaystyle tan(2x - 45) = 1$

    The inverse tangent of 1 is 45 or 225 degrees.

    Thus
    $\displaystyle 2x - 45 = 45 \implies x = 45$
    and
    $\displaystyle 2x - 45 = 225 \implies x = 135$

    To get the other two solutions note that
    $\displaystyle tan(45 + 360) = 1$
    and
    $\displaystyle tan(225 + 360) = 1$

    Thus
    $\displaystyle 2x - 45 = 405 \implies x = 225$
    and
    $\displaystyle 2x - 45 = 585 \implies x = 315$

    We would do this again, eg tan(45 + 2 * 360) = 1, but the solutions for x come out to be greater than 360, so we are done.

    -Dan
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  3. #3
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    Thankyou!!!!!!!!!!!!!!!! Thats great.
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  4. #4
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    Sorry to be annoying, but how do you get 225degrees
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by orangesmarties View Post
    Sorry to be annoying, but how do you get 225degrees
    The reference angle for tan(x) = 1 is x = 45. Tangent is positive in QI and QIII. So the possible values for x between 0 and 360 are 45 and 180 + 45 = 225.

    -Dan
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  6. #6
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    Thanks again Dan- thats been a great help!!
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  7. #7
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    Hello, orangesmarties!

    Solve: .$\displaystyle \tan(2x-45^o) \:=\;1,\quad 0 \leq x \leq 360^o$

    Answers: .$\displaystyle x \;= \;45^o,\:135^o,\:225^o,\:315^o$
    We have: .$\displaystyle \tan(2x-45^o) \:=\:1$

    Then: .$\displaystyle 2x-45^o \;=\;45^o + 180^on$ . for some integer $\displaystyle n.$

    And: .$\displaystyle 2x \;=\;90^o + 180^on$

    . . Hence: .$\displaystyle x \;=\;45^o + 90^on$


    Let $\displaystyle n \:=\:0,1,2,3$

    . . and we have: .$\displaystyle \boxed{x \;=\;45^o,\:135^o,\:225^o,\:315^o}$

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  8. #8
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    Thanks Soroban!
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