Trigonometry

Given that sin^2x = 4/13, what is the cos^2x?
I am so lost. I need to show my work. Please help.
THANK YOU!

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  1. if sin^2 x = 4/13
    sin x = ± 2/√13
    telling me that x could in any of the 4 quadrants

    using the first quadrant triangle
    x^2 + 2^2 = √13^2
    x^2 = 9
    cos^2 = x^2/r^2 = 9/13

    or, in a shorter way:

    we know : sin^2 x + cos^2 x = 1
    cos^2 x + 4/13 = 1
    cos^2 x = 1-4/13 = 9/13

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