What is the general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi ?

The general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(11pi)/6 ,θ=(5pi)/6 ,θ=(7pi)/6

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4cos^2(θ)-3=0,0<= θ<= 2pi

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

θ = \frac{π}{6}, θ = \frac{11 π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

Solution steps

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

Solve by substitution

cos (θ) = \frac{3}{2}, cos (θ) = - \frac{3}{2}

cos (θ) = \frac{3}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{11 π}{6}

cos (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π : θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{11 π}{6}, θ = \frac{5 π}{6}, θ = \frac{7 π}{6}

2 sin (θ) - 3 = - csc (θ)

cot (θ) = - 3

2 cos^{2} (t) + 3 cos (t) = - 1

x, cos (x - y) = 1

x, cos (x - y) = 0

What is the general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi ?
The general solution for 4cos^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(11pi)/6 ,θ=(5pi)/6 ,θ=(7pi)/6