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    Just now I am doing trigonometric equations like

    cos2x - sinx = 0

    the formula I am supposed to use is 1 - 2sin^2x

    can anyone do that explaining it fully, especially the bit with the cast diagram, how do I know how many values and what ones to find out?
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    Take:
    cos2x=1-sin^2x

    And put it into the equation:
    cos2x-sinx=0

    to get:
    1-2sin^2-sinx=0

    Rearrange to get a nice(ish) quadratic:
    2sin^2+sinx-1=0

    And factorise, like any quadratic:
    (2sinx-1)(sinx+1)=0

    And I'm sure you know how to get sinx:
    sinx = \frac{1}{2}

    sinx = -1

    And solve for x.
    Hope it helped.
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    cos(a+b) = cosacosb - sinasinb

    Writing cos2x as cos(x+x) = cos^2 x - sin^2 x [where cos^2 x = (cosx)^2 etc]

    Using cos^2 x + sin^2 x = 1 you can replace cos^2 x and sin^2 x in the above expression.

    i.e. 1 - 2sin^2 x

    So your equation becomes (1 - 2sin^2 x) - sinx = 0
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    Thanks for your help I understand it much better now, but one last thing. How do I use the CAST diagram (cos all sin tan) to work out which values of x to find?

    I have 30 and 270 but the answer also has 150
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    I find the best way is to sketch the graph on sinx between 0 and 2pi

    Remember your calculator is only giving you one value for arcsin ....whatever you put. If you draw a line through the graph where x = 1/2 (one of the answers for what sinx equals) you can see it intersects at pi/6 and pi-pi/6, or 30 and 180 - 30

    Doing the same for when x=-1, you see if only intersects once in this region, at 270.
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    Oh right, I think I get you now, thanks for the help
 
 
 
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Updated: January 20, 2010
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