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

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

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4sin^2(θ)-3=0,0<= θ<= 2pi

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

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Solution steps

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

Solve by substitution

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

sin (θ) = \frac{3}{2}, 0 \leq θ \leq 2 π : θ = \frac{π}{3}, θ = \frac{2 π}{3}

sin (θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π : θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

4 sin (B) + 8 = 0

4 cos (x) - 5 = - 7

0 = 1 + 2 sin (\frac{x ^{2}}{2})

cos (2 x) - 5 cos (x) + 3 = 0

9 sin^{2} (x) + 18 sin (x) + 9 = 0

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