What is the general solution for 2cos^2(θ)-7cos(θ)+3=0,0<θ<2pi ?

The general solution for 2cos^2(θ)-7cos(θ)+3=0,0<θ<2pi is θ= pi/3 ,θ=(5pi)/3

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2cos^2(θ)-7cos(θ)+3=0,0<θ<2pi

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 < θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

2 cos^{2} (θ) - 7 cos (θ) + 3 = 0, 0 < θ < 2 π

Solve by substitution

cos (θ) = 3, cos (θ) = \frac{1}{2}

cos (θ) = 3, 0 < θ < 2 π :

No Solution

cos (θ) = \frac{1}{2}, 0 < θ < 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{5 π}{3}

csc (θ) = \frac{3}{2}

12 cos^{2} (x) + 5 sin (x) - 9 = 0

2 + 2 cos (θ) = 2

1 - sec^{2} (x) = 0

3 cot^{2} (y - \frac{π}{4}) = 1

What is the general solution for 2cos^2(θ)-7cos(θ)+3=0,0<θ<2pi ?
The general solution for 2cos^2(θ)-7cos(θ)+3=0,0<θ<2pi is θ= pi/3 ,θ=(5pi)/3