Question

Solve: 2cos^2(theta) -3cos (theta) +1 =0
0≤ theta <2pi

please show work, I'm really confused!

Answer 1

Factor the left side.

Answer 2

Then set each factor equal to 0 and solve each equation.

Answer 3

so cos(theta) (2cos - 3 ) +1 =0?

Answer 4

Factor this:
\[2x^2-3x+1=0\]

Answer 5

(x-1) (2x-1)

Answer 6

(x-1) (2x-1)=0

Answer 7

so x =1 and 1/2

Answer 8

Now set each factor equal to 0 and solve each equation.

Answer 9

But remember: x is really cos (theta) so replace x with cos (theta) and find theta

Answer 10

\[\cos \theta =1\]

Answer 11

\[\cos \theta =\frac{1}{2}\]

Answer 12

how do you find theta.. do you use the unit circle?

Answer 13

yes

Answer 14

so cosθ=1 is 0 and cosθ=1/2 is √3 /2 ?

Answer 15

At 0 degrees the x value (which is the cosine) is 1. So theta = 0 degrees
At 60 degrees the x value is 1/2 so theta = 60 degrees
At 300 degrees the x value is also 1/2 so theta is also 300 degrees.

Answer 16

ohh so those are the three values for x correct?

Answer 17

i just have to convert to rad form right?

Answer 18

No. They are the values of theta. You are trying to find the angles that have the given cosines.

Answer 19

yes. 0 degrees is also 0 radians
60 degrees is pi/3 radians
300 degrees is 5pi/3 radians

Answer 20

oh okay i understand it a lot better, thank you :)

Answer 21

yw