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posted by Argelia Wednesday, February 13, 2013 at 12:44am.
solve for all values in [0,2pie) 2cos^2x-5sinx+1=0
2(1-sin^2)-5sin+1=0 2sin^2 + 5sin - 3 = 0 (2sin-1)(sin+3) = 0 so, sinx = 1/2 or -3 -3 is not allowed, so sinx = 1/2: x = pi/6 or 5pi/6
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