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

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

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sin(x-pi)=(-sqrt(2))/2

sin (x - π) = \frac{- 2}{2}

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

Degrees

x = 40 5^{\circ} + 36 0^{\circ} n, x = 49 5^{\circ} + 36 0^{\circ} n

Solution steps

sin (x - π) = \frac{- 2}{2}

Apply the fraction rule:

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

sin (x - π) = - \frac{2}{2}

General solutions for

sin (x - 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}

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

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

Solve

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

x - π = \frac{5 π}{4} + 2 πn

Move

π

to the right side

x - π = \frac{5 π}{4} + 2 πn

Add

π

to both sides

x - π + π = \frac{5 π}{4} + 2 πn + π

Simplify

x = \frac{5 π}{4} + 2 πn + π

x = \frac{5 π}{4} + 2 πn + π

Solve

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

x - π = \frac{7 π}{4} + 2 πn

Move

π

to the right side

x - π = \frac{7 π}{4} + 2 πn

Add

π

to both sides

x - π + π = \frac{7 π}{4} + 2 πn + π

Simplify

x = \frac{7 π}{4} + 2 πn + π

x = \frac{7 π}{4} + 2 πn + π

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

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

arccos (x) = 18 0^{\circ}

sin (a) = \frac{29.7}{54.5}

1.0003 sin (3 0^{\circ}) = 1.46 sin (x)

3 tan (x) + 2 cos (x) = 0

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