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

The general solution for (1+sin(t))/2 =0 is t=(3pi)/2+2pin

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(1+sin(t))/2 =0

\frac{1 + sin ( t )}{2} = 0

t = \frac{3 π}{2} + 2 πn

Degrees

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

Solution steps

\frac{1 + sin ( t )}{2} = 0

Multiply both sides by

2

\frac{1 + sin ( t )}{2} = 0

Multiply both sides by

2

\frac{2 ( 1 + sin ( t ) )}{2} = 0 \cdot 2

Simplify

1 + sin (t) = 0

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

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{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

sin (2 θ) = 2 cos (θ)

9 cos^{2} (θ) - 12 cos (θ) = - 4

- 3 sec (\frac{θ}{3}) = - 32

4 cos (2 x - π) - 1 = 1

csc (x) = 4 sin (x)

What is the general solution for (1+sin(t))/2 =0 ?
The general solution for (1+sin(t))/2 =0 is t=(3pi)/2+2pin