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

The general solution for sin(2t)=1 is t= pi/4+pin

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sin(2t)=1

sin (2 t) = 1

t = \frac{π}{4} + πn

Degrees

t = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (2 t) = 1

General solutions for

sin (2 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}

2 t = \frac{π}{2} + 2 πn

2 t = \frac{π}{2} + 2 πn

Solve

2 t = \frac{π}{2} + 2 πn : t = \frac{π}{4} + πn

2 t = \frac{π}{2} + 2 πn

Divide both sides by

2

2 t = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

t = \frac{π}{4} + πn

t = \frac{π}{4} + πn

t = \frac{π}{4} + πn

t = \frac{π}{4} + πn

sin^{4} (x) = sin^{2} (x)

2 sin^{2} (x) - 3 sin (x) = 0

0.32 = 0.59406 cos (x)

x, 4 cos (x) = 0

sec (3 x) = - \frac{2 3}{3}

What is the general solution for sin(2t)=1 ?
The general solution for sin(2t)=1 is t= pi/4+pin