What is the general solution for sin(θ-pi)=-1/(sqrt(2)) ?

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

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sin(θ-pi)=-1/(sqrt(2))

sin (θ - π) = - \frac{1}{2}

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

Radians

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

Solution steps

sin (θ - π) = - \frac{1}{2}

Simplify

- \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}

sin (θ - π) = - \frac{2}{2}

General solutions for

sin (θ - π) = - \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 - π = \frac{5 π}{4} + 2 πn, θ - π = \frac{7 π}{4} + 2 πn

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

Solve

θ - π = \frac{5 π}{4} + 2 πn : θ = \frac{5 π}{4} + 2 πn + π

θ - π = \frac{5 π}{4} + 2 πn

Move

π

to the right side

θ - π = \frac{5 π}{4} + 2 πn

Add

π

to both sides

θ - π + π = \frac{5 π}{4} + 2 πn + π

Simplify

θ = \frac{5 π}{4} + 2 πn + π

θ = \frac{5 π}{4} + 2 πn + π

Solve

θ - π = \frac{7 π}{4} + 2 πn : θ = \frac{7 π}{4} + 2 πn + π

θ - π = \frac{7 π}{4} + 2 πn

Move

π

to the right side

θ - π = \frac{7 π}{4} + 2 πn

Add

π

to both sides

θ - π + π = \frac{7 π}{4} + 2 πn + π

Simplify

θ = \frac{7 π}{4} + 2 πn + π

θ = \frac{7 π}{4} + 2 πn + π

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

7 - 2 sin (5 (x + \frac{π}{8})) = 9

x, 4 cos^{2} (x) = 3

sin (2 x) + \frac{2}{2} = 0

cos (θ) - sin (θ) = 2

3 sin (2 x + 30) = tan (2 x + 30)

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