What is the general solution for 1=-sin(x) ?

The general solution for 1=-sin(x) is x=(3pi)/2+2pin

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1=-sin(x)

1 = - sin (x)

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

1 = - sin (x)

Switch sides

- sin (x) = 1

Divide both sides by

- 1

- sin (x) = 1

Divide both sides by

- 1

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

Simplify

sin (x) = - 1

sin (x) = - 1

General solutions for

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

cos (4 5^{\circ} - x) = \frac{1}{2}

sin (x) - sin^{2} (x) = \frac{1}{4}

20 sin (x) = 20 cos (x)

\frac{- 5 cot ( x )}{2} + \frac{7}{3} = - \frac{1}{6}

8.59 tan^{2} (θ) - 24 tan (θ) + 15.29 = 0

What is the general solution for 1=-sin(x) ?
The general solution for 1=-sin(x) is x=(3pi)/2+2pin