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 < 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

Degrees

θ = 4 5^{\circ}, θ = 13 5^{\circ}

Solution steps

sin (θ) = \frac{2}{2}, 0 \leq θ < 2 π

General solutions for

sin (θ) = \frac{2}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

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

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{4}, θ = \frac{3 π}{4}

x, 10 (cos (x))^{2} + 5 cos (x) - 5 = 0

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

cos (x) = \frac{4}{14}

\frac{sin ( 57.2 2 ^{\circ} )}{31.5} = \frac{sin ( θ )}{24.4}

tan^{2} (b) sin^{2} (b) = tan^{2} (b) + sin^{2} (b)

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