Solve 2 sin^2 θ + cos θ = 1?
2 sin^2 θ + cos θ = 1
I attempted to solve this by factoring. I started with
2(1-cos^2)+cos=1
2-2cos^2+cos = 1
-2cos^2+cos+1 = 0
2cos^2 - cos - 1 = 0
(2 cos + 1)(cos - 1)=0
cos = - 1/2, cos = 0
theta = 2pi/3 theta = pi/3
I put these answers in and they were incorrect. I'm wondering where I went wrong.
Update:Mike G, I put those answers in +2kpi and I got it all right. I kind of feel like an idiot for getting this wrong, because I was so close and made such a simple mistake, but hey, all that matters is the points.
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Correct up to
(2 cos + 1)(cos - 1)=0
cos t = -1/2 and 1
t = 2pi/3, 4pi/3 , 0, 2pi
Let's see:
2 * sin(t)^2 + cos(t) = 1
2 * (1 - cos(t)^2) + cos(t) = 1
2 - 2cos(t)^2 + cos(t) = 1
-2cos(t)^2 + cos(t) + 2 - 1 = 0
-2cos(t)^2 + cos(t) + 1 = 0
2cos(t)^2 - cos(t) - 1 = 0
cos(t) = (1 +/- sqrt(1 - 4 * 2 * (-1))) / (2 * 2)
cos(t) = (1 +/- sqrt(1 + 8)) / 4
cos(t) = (1 +/- sqrt(9)) / 4
cos(t) = (1 +/- 3) / 4
cos(t) = -2/4 , 4/4
cos(t) = -1/2 , 1
cos(t) = -1/2
t = 2pi/3 , 4pi/3
cos(t) = 1
t = 0
It looks like you did everything fine up to this point
cos = -1/2 , cos = 0
That should be cos = 1, not cos = 0
Then you said that t = pi/3, which isn't right, and you neglected the answer for cos = 1, which is t = 0.
So your algebra was spot on. It was your application of the unit circle that wasn't.
2(1-cos^2 θ)+cosθ=1
2(1-x^2)+x=1
2-2x^2+x=1
1-2x^2+x=0
2x^2-x-1=0
1+-(1+8)^0.5=1+-3=4 or -2(sorry for bad latex)
divide by 4=1 or -0.5
cos θ=1 or -0.5
x=0 or 2pi x=2pi/3 or 4pi/3 side working:cos 60 degrees=0.5 180-60=120 degrees(gives the negative)=1/3 or 2pi=2/3pi 2pi-(2pi/3)=4/3pi
Your mistake was that if cos θ-1=0 cos θ=1