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=(5pi)/4 ,θ=(7pi)/4

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2sin(θ)=-sqrt(2),0<= θ<2pi

2 sin (θ) = - 2, 0 \leq θ < 2 π

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Degrees

θ = 22 5^{\circ}, θ = 31 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{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

y, x = arcsin (y^{2})

cos^{2} (2 x) + cos (2 x) - \frac{3}{4} = 0

sin^{2} (\frac{x}{2}) = a

cos (θ) = \frac{11}{5}

10 cos (θ) = 4 sin^{2} (θ) + 2

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=(5pi)/4 ,θ=(7pi)/4