2cos^2theta-3costheta+1=0
inerval [0,2Pi)
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To solve the equation 2cos^2(theta) - 3cos(theta) + 1 = 0 and find the values of theta in the interval [0, 2Pi), you can use the quadratic formula or factor the equation. Let's factor the equation in this case:
2cos^2(theta) - 3cos(theta) + 1 = 0
We can rewrite the equation as:
(2cos(theta) - 1)(cos(theta) - 1) = 0
Now, set each factor equal to zero and solve for theta:
2cos(theta) - 1 = 0 --> cos(theta) = 1/2 --> theta = π/3, 5π/3
cos(theta) - 1 = 0 --> cos(theta) = 1 --> theta = 0, 2π
So, the values of theta that satisfy the equation in the interval [0, 2π) are:
θ = 0, π/3, 5π/3, 2π.