2sin^2x+7sinx+3=0 Find all solutions of the equation in the interval [0,2pi) Write your answer in radians in terms of pi?
回答 (3)
2016-11-26 8:25 pm
2sin^2x+7sinx+3=0
2sin^2x+6sinx+sin x + 3=0
(sin x + 3)(2sin x + 1) = 0
roots are sin x = -3, -1/2
Since sin x = -3 is impossible then sin x = -1/2 only root
and x = 2pi-pi/6 , pi + pi/6
so roots are 11pi/6 , 7pi/6
2sin^2x+6sinx+sin x + 3=0
(sin x + 3)(2sin x + 1) = 0
roots are sin x = -3, -1/2
Since sin x = -3 is impossible then sin x = -1/2 only root
and x = 2pi-pi/6 , pi + pi/6
so roots are 11pi/6 , 7pi/6
2016-11-26 8:22 pm
2 * sin(x)^2 + 7 * sin(x) + 3 = 0
sin(x) = (-7 +/- sqrt(49 - 4 * 2 * 3)) / (2 * 2)
sin(x) = (-7 +/- sqrt(49 - 24)) / 4
sin(x) = (-7 +/- sqrt(25)) / 4
sin(x) = (-7 +/- 5) / 4
sin(x) = -12/4 , -2/4
sin(x) = -3 , -1/2
sin(x) = -1/2
x = 7pi/6 , 11pi/6
sin(x) = (-7 +/- sqrt(49 - 4 * 2 * 3)) / (2 * 2)
sin(x) = (-7 +/- sqrt(49 - 24)) / 4
sin(x) = (-7 +/- sqrt(25)) / 4
sin(x) = (-7 +/- 5) / 4
sin(x) = -12/4 , -2/4
sin(x) = -3 , -1/2
sin(x) = -1/2
x = 7pi/6 , 11pi/6
2016-11-26 9:05 pm
[ 2 sin x + 1 ] [ sin x + 3 ] = 0
sin x = - 1/2
x = 4π/3 , 5π/3
sin x = - 1/2
x = 4π/3 , 5π/3
收錄日期: 2021-04-23 23:52:32
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