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

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

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2sin(θ)=sqrt(2),0<= θ<2pi

2 sin (θ) = 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

2 sin (θ) = 2, 0 \leq θ < 2 π

Divide both sides by

2

2 sin (θ) = 2

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{2}{2}

Simplify

sin (θ) = \frac{2}{2}

sin (θ) = \frac{2}{2}

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}

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

cos (t) - cos (2 t) = - 0.75

2 cos (A) = - 3

4 sin (x) cos (x) = cos (x)

cos (x) = 2 sin (x) cos (x)

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