What is the general solution for sqrt(2)sin(θ)+1=0 ?

The general solution for sqrt(2)sin(θ)+1=0 is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin

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sqrt(2)sin(θ)+1=0

2 sin (θ) + 1 = 0

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

Degrees

θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) + 1 = 0

Move

1

to the right side

2 sin (θ) + 1 = 0

Subtract

1

from both sides

2 sin (θ) + 1 - 1 = 0 - 1

Simplify

2 sin (θ) = - 1

2 sin (θ) = - 1

Divide both sides by

2

2 sin (θ) = - 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{- 1}{2}

Simplify

\frac{2 sin ( θ )}{2} = \frac{- 1}{2}

Simplify

\frac{2 sin ( θ )}{2} : sin (θ)

\frac{2 sin ( θ )}{2}

Cancel the common factor:

2

= sin (θ)

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (θ) = - \frac{2}{2}

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

sin (t) = 0

sin (θ) = 2

sin (x) = \frac{12}{13}

cos (θ) = \frac{3}{5}

sin (x) = - 2

What is the general solution for sqrt(2)sin(θ)+1=0 ?
The general solution for sqrt(2)sin(θ)+1=0 is θ=(5pi)/4+2pin,θ=(7pi)/4+2pin