Asked by Argelia
solve for all values in [0,2pie)
2cos^2x-5sinx+1=0
2cos^2x-5sinx+1=0
Answers
Answered by
Steve
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
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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