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 \leq θ \leq 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

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

Solve by substitution

cos (θ) = 3, cos (θ) = \frac{1}{2}

cos (θ) = 3, 0 \leq θ \leq 2 π :

No Solution

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

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{5 π}{3}

4 sin (x) cos (x) + 2 sin (x) - 2 cos (x) = 1

sin (x) cos^{2} (x) = sin (x)

3 sin (x) cos (x) = cos^{2} (x)

2 cos^{2} (4 x) + 1 = 0

1 + cos (2 θ) = - 3 sin (θ)

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