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    Solve the following equation for  \theta , in the interval  0<\theta<360^0 :

     (Sin \theta - 1)(5cos \theta + 3) = 0

    Why can I not do them individually? So...

     Sin \theta - 1 = 0   5cos \theta + 3 = 0

     Sin \theta = 1  cos \theta = -\frac {3}{5}

    then proceed from there?
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    (Original post by Naruke)
    Solve the following equation for  \theta , in the interval  0<\theta<360^0 :

     (Sin \theta - 1)(5cos \theta + 3) = 0

    Why can I not do them individually? So...

     Sin \theta - 1 = 0   5cos \theta + 3 = 0

     Sin \theta = 1  cos \theta = -\frac {3}{5}

    then proceed from there?
    correct
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    (Original post by TeeEm)
    correct
    My answers don't correspond with the mark scheme
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    (Original post by Naruke)
    My answers don't correspond with the mark scheme
    then you must be doing something else wrong
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    I'm sorry. I'm a bit retarded. I was looking at the wrong mark scheme
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    (Original post by TeeEm)
    then you must be doing something else wrong
    Sorry. Another question, same range.

     tan \theta = tan \theta (2+3sin \theta)

     \frac{tan \theta}{tan \theta} = 2 + 3sin \theta

     Sin \theta = -\frac {1}{3}

    With this I got 2 of the solutions. However, apparently 180^o & 360^o are solutions to?
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    Anyone?
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    (Original post by Naruke)
    Anyone?
    Answered. :yep:
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    (Original post by Naruke)
    Anyone?
    you cannot divide by tangent
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    (Original post by Naruke)
    Sorry. Another question, same range.

     tan \theta = tan \theta (2+3sin \theta)

     \frac{tan \theta}{tan \theta} = 2 + 3sin \theta

     Sin \theta = -\frac {1}{3}

    With this I got 2 of the solutions. However, apparently 180^o & 360^o are solutions to?
    This guy is some level of flop.
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    (Original post by Username002.5)
    This guy is some level of flop.
    I know, but all the embarrassing questions will pay off when I get an A.
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    (Original post by Naruke)
    I know, but all the embarrassing questions will pay off when I get an A.
    It's not embarrassing in the least nor are you a flop in anyway.
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    (Original post by Zacken)
    It's not embarrassing in the least nor are you a flop in anyway.

    Cheers!

    This will probably be my last question on trig:

     2tan^2x - tan x = 6
     tan x (2tan x - 1) = 6
     tan x = 6 or  2tan x - 1 = 6

    Why would this approach be wrong?
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    (Original post by Naruke)
    Cheers!

    This will probably be my last question on trig:

     2tan^2x - tan x = 6
     tan x (2tan x - 1) = 6
     tan x = 6 or  2tan x - 1 = 6

    Why would this approach be wrong?
    You can only factorise stuff and equal it to 0 if it's an equation of them form =0.

    E.g, if you had x^2 - 2x = -1

    Would you collect that as a quadratic and aolve or would you do x(x-2)= -1 so x = -1 and x -2 = -1.

    Does that sound reasonable? What are the correct answers to that quadratic?
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    (Original post by Naruke)
    Cheers!

    This will probably be my last question on trig:

     2tan^2x - tan x = 6
     tan x (2tan x - 1) = 6
     tan x = 6 or  2tan x - 1 = 6

    Why would this approach be wrong?
    You can treat this as an ordinary quadratic by taking 6 away from both sides
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    (Original post by Zacken)
    You can only factorise stuff and equal it to 0 if it's an equation of them form =0.

    E.g, if you had x^2 - 2x = -1

    Would you collect that as a quadratic and aolve or would you do x(x-2)= -1 so x = -1 and x -2 = -1.

    Does that sound reasonable? What are the correct answers to that quadratic?
    Oh, yes. I think I understand it now. I was looking at these trig equations all wrong. I got -116.6, -56.3, 63.4, 123.7 all to 1 dp which corresponds with the answers.

    Thanks again!
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    (Original post by Naruke)
    Oh, yes. I think I understand it now. I was looking at these trig equations all wrong. I got -116.6, -56.3, 63.4, 123.7 all to 1 dp which corresponds with the answers.

    Thanks again!
    No problem! Keep your threads and questions coming. I'm predicting a good grade for you come summer.
 
 
 
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