What is the general solution for sin(θ)=(sqrt(2))/2 ,0<= θ<= 2pi ?

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

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sin(θ)=(sqrt(2))/2 ,0<= θ<= 2pi

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

θ = \frac{π}{4}, θ = \frac{3 π}{4}

Solution steps

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

General solutions for

sin (θ) = \frac{2}{2}

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

x, cos (x) = \frac{2}{3}

3 sin (x) + 2 = 1.5

3 sec^{2} (x) - 2 tan (x) - 4 = 0

9 sin^{2} (x) - 1 = 0

sin (x) = \frac{1}{3.5}

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