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  1. #1
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    Trig question

    Solve the equation sin2A -cos^2A=0
    giving values of A between 0 degrees and 360 degrees.

    THANKS!
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  2. #2
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    Quote Originally Posted by Oasis1993 View Post
    Solve the equation sin2A -cos^2A=0
    giving values of A between 0 degrees and 360 degrees.

    THANKS!
    Substitute sin(2A) = 2 sin(A) cos(A). Then factorise and use the null factor law to get two simple trig equatins to solve.
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    I didnt understand. How am i supposed to factorise it?
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    Quote Originally Posted by Oasis1993 View Post
    Solve the equation sin2A -cos^2A=0

    Quote Originally Posted by Oasis1993 View Post

    giving values of A between 0 degrees and 360 degrees.

    THANKS!

    hi all
    my solution

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  5. #5
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    Quote Originally Posted by Oasis1993 View Post
    I didnt understand. How am i supposed to factorise it?
    Did you substitute sin(2A) = 2 sin(A) cos(A) like I said to? If so, you should see how to factorise by taking out the common factor of cos(A).
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