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    Jan 2007 Core2 Paper

    "Find all the solutions, in the interval 0 <= x < 2pi, of the equation

    2cos^2(x) + 1 = 5 sin(x)

    giving each solution in terms of pi"

    I've done the whole C2 paper except this question, and gone back to it a few times.

    I'm just requesting a full workout and method for this so if I come across another one I will know how to tackle it, thanks

    Jorge x
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    I can't do Latex too well, so it'd be really ugly if i tried, so i'll just say, you can swap cos^2 for 1-sin^2, and you've got a quadratic for sin. They should all go something like that.
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    I've tried, the question is just not working for me
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    (Original post by Jorgeyy)
    I've tried, the question is just not working for me
    When you replaced cos^2 with 1-sin^2 what quadratic equation did you get?
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    (Original post by vc94)
    When you replaced cos^2 with 1-sin^2 what quadratic equation did you get?
    I'll find my workings in my book later,

    but trust me, ive left this alone for 24h come back to it and im still stuck

    I would appreciate if someone could organise the answer in steps so I can see how to tackle these questions
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    (Original post by Jorgeyy)
    Jan 2007 Core2 Paper

    "Find all the solutions, in the interval 0 <= x < 2pi, of the equation

    2cos^2(x) + 1 = 5 sin(x)

    giving each solution in terms of pi"

    (Original post by Jorgeyy)
    I would appreciate if someone could organise the answer in steps so I can see how to tackle these questions
    Right.

    2cos^2x + 1 = 5sinx
    (You MUST remember this identity : cos^2x + sin^2x = 1 which implies that cos^2x = 1 - sin^2x )
    Substitute this into your equation (pretty much like solving a simultaneous equation)

    cos^2x=1-sin^2x

    2 (1 - sin^2x) + 1 = 5sinx

    2 - 2sin^2x + 1 = 5sinx

    rearrange to get it in the form of a quadratic.

    2sin^2x + 5sinx - 3 = 0
    It might look scary but it's exactly like a quadratic you solved in Core 1.

    TO make it more familiar i'll replace sinx with y .

    so 2y^2 + 5y - 3 =0

    factorising : (2y-1) ( y+3)=0

    Since sinx has a maximum value of 1 and a minimum value of -1 the solution of y=-3 would be rejected (radians or degrees, doesnt matter)

    so our solution is that 2y-1 = 0
    therefore y = 1/2

    so back to our question sinx = 1/2

    to find your first solution just do sin^-1 on your calculator (radian mode as it asks in terms of pi)

    SO our first solution is sin^-1(1/2) = pi/6

    The second solution you would use the CAST method and find that sin is positive in the first two quadrants only.

    So the second solution is 180 (pi) - pi/6 which = 5pi/6

    And that's all. pretty standard question, just make sure you know the identity .

    Also if in doubt substitute your solutions into the starting equation and it should work ( if not go back and check your solution)
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    (Original post by sohail.s)
    Right.

    2cos^2x + 1 = 5sinx
    (You MUST remember this identity : cos^2x + sin^2x = 1 which implies that cos^2x = 1 - sin^2x )
    Substitute this into your equation (pretty much like solving a simultaneous equation)

    cos^2x=1-sin^2x

    2 (1 - sin^2x) + 1 = 5sinx

    2 - 2sin^2x + 1 = 5sinx

    rearrange to get it in the form of a quadratic.

    2sin^2x + 5sinx - 3 = 0
    It might look scary but it's exactly like a quadratic you solved in Core 1.

    TO make it more familiar ill replace sinx with y .

    so 2y^2 + 5y - 3 =0

    factorising : (2y-1) ( y+3)=0

    Since sinx has a maximum value of 1 and a minimum value of -1 the solution of y=-3 would be rejected (radians or degrees, doesnt matter)

    so our solution is that 2y-1 = 0
    therefore y = 1/2

    so back to our question sinx = 1/2

    to find your first solution just do sin^-1 on your calculator (radian mode as it asks in terms of pi)

    SO our first solution is sin^-1(1/2) = pi/6

    The second solution you would use the CAST method and find that sin is positive in the first two quadrants only.

    So the second solution is 180 (pi) - pi/6 which = 5pi/6

    And that's all. pretty standard question, just make sure you know the identity .

    Also if in doubt substitute your solutions into the starting equation and it should work ( if not go back and check your solution)
    I love you.

    I forgot the limits to the Sin and Cos graphs of -1 to 1 and that messed me over lol !

    Also, is it ok to use the formula?

    Sin being... 180 - angle1
    Cos being... -angle1
    Tan being... angle1 + 180

    I should really learn the CAST method properly as well but i'm more of a formula person
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    (Original post by Jorgeyy)
    I love you.

    I forgot the limits to the Sin and Cos graphs of -1 to 1 and that messed me over lol !

    Also, is it ok to use the formula?

    Sin being... 180 - angle1
    Cos being... -angle1
    Tan being... angle1 + 180

    I should really learn the CAST method properly as well but i'm more of a formula person
    yeah thts fine tbh, all of them originate from the cast method tbh

    so sin = 180 - angle1
    cos = 360 - angle1
    tan = 180 + angle 1
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    I got taught Cos was -angle1 (+n360) so same thing I guess just teachers preference :P

    Thanks again man, we've just started covering trig in class its our last core2 topic so hopefully I'll understand it lightning speed ready for the exam!

    Big rep x
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    (Original post by Jorgeyy)
    I got taught Cos was -angle1 (+n360) so same thing I guess just teachers preference :P

    Thanks again man, we've just started covering trig in class its our last core2 topic so hopefully I'll understand it lightning speed ready for the exam!

    Big rep x
    Is Core 2 part of A level Maths or F. Maths?
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    (Original post by Americaniamh)
    Is Core 2 part of A level Maths or F. Maths?
    Core1 + Core2 + a side module such as stats or mechanics etc makes up AS Maths
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    (Original post by Jorgeyy)
    Core1 + Core2 + a side module such as stats or mechanics etc makes up AS Maths
    Thanks!
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    2cos^2x+1=5sinx
    Replace cos^2x for 1-sin^2x
    2(1-sin^2x)+1=5sinx
    2-2sin^2x+1=5sinx
    0=2sin^2x+5sinx-3
    Replace sinx with y
    2y^2+5y-3=0
    Factorise
    (2y-1)(y+3)
    Sinx=1/2 sin-1x=30
    I can get this far but don't understand how to give your answers in terms of pi...if anyone can help
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    (Original post by Tink.0818)
    2cos^2x+1=5sinx
    Replace cos^2x for 1-sin^2x
    2(1-sin^2x)+1=5sinx
    2-2sin^2x+1=5sinx
    0=2sin^2x+5sinx-3
    Replace sinx with y
    2y^2+5y-3=0
    Factorise
    (2y-1)(y+3)
    Sinx=1/2 sin-1x=30
    I can get this far but don't understand how to give your answers in terms of pi...if anyone can help
    jesus I thought you were answering a 4 year old question then..

    sin x = 1/2
    arcsin (1/2) = 30 degrees

    there's other answers as well, 150 for example

    to get from degrees to radians, divide by 180 then multiply by pi

    so 30 degrees in radians is 1/6pi etc
 
 
 
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