What is the general solution for 2cos^2(θ)-3cos(θ)+1=0,(3pi)/2 <θ<2pi ?

The general solution for 2cos^2(θ)-3cos(θ)+1=0,(3pi)/2 <θ<2pi is θ=(5pi)/3

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2cos^2(θ)-3cos(θ)+1=0,(3pi)/2 <θ<2pi

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, \frac{3 π}{2} < θ < 2 π

θ = \frac{5 π}{3}

Solution steps

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, \frac{3 π}{2} < θ < 2 π

Solve by substitution

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1, \frac{3 π}{2} < θ < 2 π :

No Solution

cos (θ) = \frac{1}{2}, \frac{3 π}{2} < θ < 2 π : θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{5 π}{3}

- 2 cos^{2} (x) + cos (x) = 0

3 cos (x) - 2 sin (2 x) = 0

4 cos (2 t) = 2

tan (θ) = \frac{10}{5.32}

2 cos (x) = \frac{6}{7}

What is the general solution for 2cos^2(θ)-3cos(θ)+1=0,(3pi)/2 <θ<2pi ?
The general solution for 2cos^2(θ)-3cos(θ)+1=0,(3pi)/2 <θ<2pi is θ=(5pi)/3