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    I'm currently doing C2 in A Level Maths atm but I've seen that the particular question I am struggling on is in C3?? So can someone please help me.

    1) Show that the equation

    tan 2x = 5 sinx 2x

    can be written in the form

    (1-5 cos 2x) sin 2x = 0

    Here I just rearranged the first equation using: sin 2x / cos 2x = 5 sin 2x

    2) Hence solve for 0 <= x <= 180

    tan 2x = 5 sin 2x

    Answer to 1dp where appropriate. You must show clearly how answers are obtained


    I have attempted this in too many ways and I am just confusing myself. Please Help
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    (Original post by basketballgirl11)
    I'm currently doing C2 in A Level Maths atm but I've seen that the particular question I am struggling on is in C3?? So can someone please help me.

    1) Show that the equation

    tan 2x = 5 sinx 2x

    can be written in the form

    (1-5 cos 2x) sin 2x = 0

    Here I just rearranged the first equation using: sin 2x / cos 2x = 5 sin 2x

    2) Hence solve for 0 <= x <= 180

    tan 2x = 5 sin 2x

    Answer to 1dp where appropriate. You must show clearly how answers are obtained


    I have attempted this in too many ways and I am just confusing myself. Please Help
    Could you post some working so we can help? The Maths forum does not encourage full solutions to be posted.
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    (Original post by Muttley79)
    Could you post some working so we can help? The Maths forum does not encourage full solutions to be posted.
    Which part do you want working out for? I haven't really got any accurate solution to the last part yet
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    (Original post by basketballgirl11)

    Here I just rearranged the first equation using: sin 2x / cos 2x = 5 sin 2x
    Now multiply both sides by \cos 2x and bring everything over to one side and factorise.
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    (Original post by Zacken)
    Now multiply both sides by \cos 2x and bring everything over to one side and factorise.
    I have done that already, but I am struggling with solving

    So i done sin2x/cos 2x = 5sin 2x

    sin 2x = 5sin2x x cos 2x

    sin 2x - 5 sin 2x x cos 2x = 0
    (1-5cos2x) sin 2x = 0

    And then would cos x = 1/5???
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    (Original post by basketballgirl11)
    I have done that already, but I am struggling with solving

    So i done sin2x/cos 2x = 5sin 2x

    sin 2x = 5sin2x x cos 2x

    sin 2x - 5 sin 2x x cos 2x = 0
    (1-5cos2x) sin x = 0

    And then would cos x = 1/5???
    What happens when you have a quadratic (x-2)(x-3) = 0? You get two solutions, right? By equating each bracket to 0?

    In thise case, you have two solution sets as well, one is generated by 1 - 5\cos 2x = 0 \Rightarrow \cos 2x = \frac{1}{5} (solve this and find all the solutions)

    The other is given by \sin 2x = 0 (solve this and find all the solutions)

    Then put all these solutions in one list x = 0, pi, etc... from both of the above solutions and that's your answer.
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    (Original post by Zacken)
    What happens when you have a quadratic (x-2)(x-3) = 0? You get two solutions, right? By equating each bracket to 0?

    In thise case, you have two solution sets as well, one is generated by 1 - 5\cos 2x = 0 \Rightarrow \cos 2x = \frac{1}{5} (solve this and find all the solutions)

    The other is given by \sin x = 0 (solve this and find all the solutions)

    Then put all these solutions in one list x = 0, pi, etc... from both of the above solutions and that's your answer.
    Okay, thank you

    I shall try it and then post my answers to double check
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    (Original post by basketballgirl11)
    Okay, thank you

    I shall try it and then post my answers to double check
    Awesome, will check them for you.
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    (Original post by Zacken)
    Awesome, will check them for you.
    Okay because it is for 0<= x <= 180

    I got for cos x

    39.3 (1dp) and 140.8 (1dp)

    And then sin x

    0 and 180

    Fingers crossed it's correct
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    (Original post by basketballgirl11)
    Okay because it is for 0<= x <= 180

    I got for cos x

    39.3 (1dp) and 140.8 (1dp)
    These are both correct.

    And then sin x

    0 and 180

    Fingers crossed it's correct
    Sorry! I wrote \sin x = 0 when I should have written \sin 2x = 0
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    (Original post by Zacken)
    These are both correct.



    Sorry! I wrote \sin x = 0 when I should have written \sin 2x = 0
    So would sin x just be 0 and 90?
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    (Original post by basketballgirl11)
    So would sin x just be 0 and 90?
    Almost! \sin 2x = 0 \Rightarrow 2x = 0, 180^{\circ}, 360^{\circ}, so x = 0, 90^{\circ}, ?.
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    (Original post by Zacken)
    Almost! \sin 2x = 0 \Rightarrow 2x = 0, 180^{\circ}, 360^{\circ}, so x = 0, 90^{\circ}, ?.
    Okay so I would get

    Sin x = 0, 90 and 180??
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    (Original post by basketballgirl11)
    Okay so I would get

    Sin x = 0, 90 and 180??
    You keep writing sin x = 0, 90, 180. But you need to write x = 0, 90, 180! (*) Otherwise, those values are correct, yes.


    (*) It's like me saying solve x^2 = 9 and saying the solutions are x^2 = 3, -3. That's obviously nonsense, the correct form is x = 3, -3.
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    (Original post by Zacken)
    You keep writing sin x = 0, 90, 180. But you need to write x = 0, 90, 180! (*) Otherwise, those values are correct, yes.


    (*) It's like me saying solve x^2 = 9 and saying the solutions are x^2 = 3, -3. That's obviously nonsense, the correct form is x = 3, -3.
    Okay thank you, I seem to write it perfectly in my writing so it's all good
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    (Original post by basketballgirl11)
    Okay thank you, I seem to write it perfectly in my writing so it's all good
    Oh, okay. I suspected as much but I just needed to make sure. Otherwise, good job!
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    (Original post by basketballgirl11)
    Okay so I would get

    Sin x = 0, 90 and 180??
    Remember you can always substitute your 'answers' into the equation to check.
 
 
 
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