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    How to go about solving trig equations such as these:Name:  IMG_20150516_212558531.jpg
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    Basically, any equation where the same function is on both sides of the equations. Im tempted to cancel it out just by doing the inverse, but i know thats incorrect and i will lose solutions. Could anyone explain the process of solving questions likr these? Thanks in advance
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    I'm pretty sure?


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    For the 2nd one you divide 48 by 2
    Then you do 180-48 which is 132, divide that by 2
    Then 360 add 48 divide 2
    And finally 360 add 180 - 48 for the final answer. You should have 4 X values.
    Don't know about the first one
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    (Original post by Isis_on_the_cake)
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    I'm pretty sure?


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    Thing i dont get with that is you do the inverse of tan, but then add 180 to the answer anyway as if it was a function of tan? You got all the right solutions, i just dont quite understand why you still add 180 when you 'cancelled out' the tan function, if that makes sense.
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    (Original post by Rk2k14)
    For the 2nd one you divide 48 by 2
    Then you do 180-48 which is 132, divide that by 2
    Then 360 add 48 divide 2
    And finally 360 add 180 - 48 for the final answer. You should have 4 X values.
    Don't know about the first one
    So you do 180-answer, as if it was a sin function but you do the inverse of sin in the first place? Thats what i cant quite get my head around
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    (Original post by mickel_w)
    Thing i dont get with that is you do the inverse of tan, but then add 180 to the answer anyway as if it was a function of tan? You got all the right solutions, i just dont quite understand why you still add 180 when you 'cancelled out' the tan function, if that makes sense.
    (Original post by mickel_w)
    So you do 180-answer, as if it was a sin function but you do the inverse of sin in the first place? Thats what i cant quite get my head around
    Because you have an equation in \theta, but theta must satisfy whatever trig function you started with -> ergo, you can use the appropriate formulas to find the other solutions.
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    (Original post by SamKeene)
    Because you have an equation in \theta, but theta must satisfy whatever trig function you started with -> ergo, you can use the appropriate formulas to find the other solutions.
    Ok, thank you. Makes sense now
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    (Original post by mickel_w)
    Ok, thank you. Makes sense now
    Note even when you have a simple equation you are used to solving like:

    \sin{\theta}=1 \ \ \ \ \ \ \ \ 0<\theta<180

    When you solve this, you do exactly you've been doing just now, you apply the inverse to both sides:

    \arcsin{\big( \sin{\theta} \big)} = \arcsin{(1)}

    \theta = \dfrac{\pi}{2}

    And then you apply the various rules to find the other solutions if needed (which it isn't in the aforementioned domain).
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    Slightly different way than how the other did it I think.

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    (Original post by mickel_w)
    How to go about solving trig equations such as these:Name:  IMG_20150516_212558531.jpg
Views: 148
Size:  138.8 KBAttachment 400053400063

    Basically, any equation where the same function is on both sides of the equations. Im tempted to cancel it out just by doing the inverse, but i know thats incorrect and i will lose solutions. Could anyone explain the process of solving questions likr these? Thanks in advance
    In cases such as these the unit circle is your best bet. A useful substitution is also handy like let v=2x.
    Note we have two cases with stuff like this in sinx=sinw
    infinite solutions are X=w+2npi or X=pi-w-2npi where n are integers
    Using these gives all solutions. Since you have restrictions on tangents use some values of n to limit the answers.


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