Math CentralQuandaries & Queries


Question from Mary, a student:

Solve the following trigonometric equation over 0 being less than or equal to A but less than 2pie

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


and then you start to solve

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

Which one is it?


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