What is the general solution for sin(5t)-1=0 ?

The general solution for sin(5t)-1=0 is t= pi/(10)+(2pin)/5

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sin(5t)-1=0

sin (5 t) - 1 = 0

t = \frac{π}{10} + \frac{2 πn}{5}

Degrees

t = 1 8^{\circ} + 7 2^{\circ} n

Solution steps

sin (5 t) - 1 = 0

Move

1

to the right side

sin (5 t) - 1 = 0

Add

1

to both sides

sin (5 t) - 1 + 1 = 0 + 1

Simplify

sin (5 t) = 1

sin (5 t) = 1

General solutions for

sin (5 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}

5 t = \frac{π}{2} + 2 πn

5 t = \frac{π}{2} + 2 πn

Solve

5 t = \frac{π}{2} + 2 πn : t = \frac{π}{10} + \frac{2 πn}{5}

5 t = \frac{π}{2} + 2 πn

Divide both sides by

5

5 t = \frac{π}{2} + 2 πn

Divide both sides by

5

\frac{5 t}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 t}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 t}{5} : t

\frac{5 t}{5}

Divide the numbers:

\frac{5}{5} = 1

= t

Simplify

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} : \frac{π}{10} + \frac{2 πn}{5}

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

\frac{\frac{π}{2}}{5} = \frac{π}{10}

\frac{\frac{π}{2}}{5}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{π}{10}

= \frac{π}{10} + \frac{2 πn}{5}

t = \frac{π}{10} + \frac{2 πn}{5}

t = \frac{π}{10} + \frac{2 πn}{5}

t = \frac{π}{10} + \frac{2 πn}{5}

t = \frac{π}{10} + \frac{2 πn}{5}

2 sin (θ) cos (θ) - sin (θ) = 0

9 sin (x) tan (x) - 2 tan (x) = 0

3 cot (3 x) + 1 = 0

tan (4 x) = 3

cos (x) + 2 = - cos (x)

What is the general solution for sin(5t)-1=0 ?
The general solution for sin(5t)-1=0 is t= pi/(10)+(2pin)/5