Mathematics, 14.04.2021 19:50 dondre54
Solve on the interval [0,2pi) :
2 sin^2 x-3 sin x + 1 = 0 x= pi, x=2pi/3, x=5pi/3
x=2pi, x=pi/4, x=5pi/4
x=2pi, x=pi/3
x=pi/2, x=pi/6, x=5pi/6
Answers: 3
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Solve on the interval [0,2pi) :
2 sin^2 x-3 sin x + 1 = 0 x= pi, x=2pi/3, x=5pi/3
x=2pi, x=pi...
x=2pi, x=pi...
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