What is the general solution for 1-sin(t)=0 ?

The general solution for 1-sin(t)=0 is t= pi/2+2pin

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1-sin(t)=0

1 - sin (t) = 0

t = \frac{π}{2} + 2 πn

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

1 - sin (t) = 0

Move

1

to the right side

1 - sin (t) = 0

Subtract

1

from both sides

1 - sin (t) - 1 = 0 - 1

Simplify

- sin (t) = - 1

- sin (t) = - 1

Divide both sides by

- 1

- sin (t) = - 1

Divide both sides by

- 1

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

Simplify

sin (t) = 1

sin (t) = 1

General solutions for

sin (t) = 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}

t = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

θ, 6 = ρ sin (φ) sin (θ)

tan^{2} (θ) sec^{2} (θ) = 3

2 cos (θ) = cos (θ)

sin^{2} (x) \cdot cos (x) + 1 = 0

sin (B) = \frac{1}{2}, a = 170

What is the general solution for 1-sin(t)=0 ?
The general solution for 1-sin(t)=0 is t= pi/2+2pin