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

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

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

sin (x) = \frac{- 2}{2}

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

Degrees

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

Solution steps

sin (x) = \frac{- 2}{2}

Apply the fraction rule:

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

sin (x) = - \frac{2}{2}

General solutions for

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

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

sin (x) = - \frac{1}{2}

cos (2 x) - 2 cos (x) = - 1

cos (x) = \frac{3}{2}

- 3 sin (3 x) = 0

cos (2 x) - cos (6 x) = 0

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