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 Question from Mary, a student: Solve the following trigonometric equation over 0 being less than or equal to A but less than 2pie 2cos(2A)=-2^1/2 I have it partially worked out 2(2cos^2A) = -2^1/2 4cos^2A = -2^1/2 16cos^4A = 2 16cos^4A - 2 = 0 2(8cos^4A - 1) =0 I dont know where to go from here and I don't know if I am on the right track. If you could help me figure this out it would be much appreciated. thanks.

Hi Mary,

You state the equation as

$2\cos(2A)=-2^{1/2}$

and then you start to solve

$2(2\cos^2A) = -2^{1/2}$

Which one is it?

Penny

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