What is the general solution for 2sin(4t)=2sin(4t)cos(4t) ?

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

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Popular Trigonometry >

2sin(4t)=2sin(4t)cos(4t)

Solution

2 sin (4 t) = 2 sin (4 t) cos (4 t)

Solution

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

Degrees

t = 0^{\circ} + 9 0^{\circ} n, t = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

2 sin (4 t) = 2 sin (4 t) cos (4 t)

Subtract

2 sin (4 t) cos (4 t)

from both sides

2 sin (4 t) - 2 sin (4 t) cos (4 t) = 0

Factor

2 sin (4 t) - 2 sin (4 t) cos (4 t) : - 2 sin (4 t) (cos (4 t) - 1)

2 sin (4 t) - 2 sin (4 t) cos (4 t)

Factor out common term

- 2 sin (4 t)

= - 2 sin (4 t) (- 1 + cos (4 t))

- 2 sin (4 t) (cos (4 t) - 1) = 0

Solving each part separately

sin (4 t) = 0 or cos (4 t) - 1 = 0

sin (4 t) = 0 : t = \frac{πn}{2}, t = \frac{π}{4} + \frac{πn}{2}

sin (4 t) = 0

General solutions for

sin (4 t) = 0

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}

4 t = 0 + 2 πn, 4 t = π + 2 πn

4 t = 0 + 2 πn, 4 t = π + 2 πn

Solve

4 t = 0 + 2 πn : t = \frac{πn}{2}

4 t = 0 + 2 πn

0 + 2 πn = 2 πn

4 t = 2 πn

Divide both sides by

4

4 t = 2 πn

Divide both sides by

4

\frac{4 t}{4} = \frac{2 πn}{4}

Simplify

t = \frac{πn}{2}

t = \frac{πn}{2}

Solve

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

4 t = π + 2 πn

Divide both sides by

4

4 t = π + 2 πn

Divide both sides by

4

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

Simplify

t = \frac{π}{4} + \frac{πn}{2}

t = \frac{π}{4} + \frac{πn}{2}

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

cos (4 t) - 1 = 0 : t = \frac{πn}{2}

cos (4 t) - 1 = 0

Move

1

to the right side

cos (4 t) - 1 = 0

Add

1

to both sides

cos (4 t) - 1 + 1 = 0 + 1

Simplify

cos (4 t) = 1

cos (4 t) = 1

General solutions for

cos (4 t) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

4 t = 0 + 2 πn

4 t = 0 + 2 πn

Solve

4 t = 0 + 2 πn : t = \frac{πn}{2}

4 t = 0 + 2 πn

0 + 2 πn = 2 πn

4 t = 2 πn

Divide both sides by

4

4 t = 2 πn

Divide both sides by

4

\frac{4 t}{4} = \frac{2 πn}{4}

Simplify

t = \frac{πn}{2}

t = \frac{πn}{2}

t = \frac{πn}{2}

Combine all the solutions

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

Graph

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Frequently Asked Questions (FAQ)

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