What is the general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi ?

The general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6

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12sin^2(θ)-3=0,0<= θ<= 2pi

12 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

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

Solution steps

12 sin^{2} (θ) - 3 = 0, 0 \leq θ \leq 2 π

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{5 π}{6}

sin (θ) = - \frac{1}{2}, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

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

0 = - 7.5 sin (\frac{π}{10} (t - 5)) + 8

cot (11 x + 1) = tan (6 x + 4)

2 sin (2 x) = cos (2 x + 3 0^{\circ})

tan (θ) = \frac{15}{25}

tan (φ) = 3

What is the general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi ?
The general solution for 12sin^2(θ)-3=0,0<= θ<= 2pi is θ= pi/6 ,θ=(5pi)/6 ,θ=(7pi)/6 ,θ=(11pi)/6