solve cos2x-5sinx-3=0?
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solve cos2x-5sinx-3=0?

[From: ] [author: ] [Date: 17-05-10] [Hit: ]
11pi/6 The general solution is x = 7pi/6 + 2pi n, where n is any integer x = 11pi/6 + 2pi n,......
u = (-(-5) +/- sqrt((-5)^2 - 4(-2)(-2))) / (2(-2))
sin(x) = (5 +/- sqrt(25 - 16)) / (-4)
sin(x) = (5 +/- sqrt(9)) / (-4)
sin(x) = (5 +/- 3) / (-4)
sin(x) = (8 or -2) / (-4)
sin(x) = -2 or 1/2
Solutions:
x = 2*pi*n - arcsin(2) radians, for any integer n
x = 2*pi*n * pi + arcsin(2) radians, for any integer n
x = 2*pi*n + (pi/6) radians, for any integer n
x = 2*pi*n + (5*pi/6) radians, for any integer n
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cidyah say: cos 2x = 1-2 sin^2 x
cos 2x - 5 sin x - 3 = 0
1-2sin^2x - 5 sin x - 3 = 0
-2sin^2 x - 5 sin x - 2 = 0
2sin^2 x + 5 sin x + 2 = 0

Let y=sin x
2y^2+5y+2=0
2y^2+4y+y+2=0
2y(y+2)+1(y+2)=0
(y+2)(2y+1)=0
y+2 = 0
y=-2
sin x = -2 (no solution)

2y+1=0
2y=-1
y=-1/2
sin x = -1/2
x = 7pi/6, 11pi/6

The general solution is
x = 7pi/6 + 2pi n, where n is any integer
x = 11pi/6 + 2pi n, where n is any integer
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keywords: solve,cos,sinx,solve cos2x-5sinx-3=0?
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