What is the general solution for 4cos(2t)+1=3,0, pi/2 ?

The general solution for 4cos(2t)+1=3,0, pi/2 is No Solution for t\in\mathbb{R}

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4cos(2t)+1=3,0, pi/2

Solution

4 cos (2 t) + 1 = 3, 0, \frac{π}{2}

Solution

No Solution for t \in R

Solution steps

4 cos (2 t) + 1 = 3, 0, \frac{π}{2}

Move

1

to the right side

4 cos (2 t) + 1 = 3

Subtract

1

from both sides

4 cos (2 t) + 1 - 1 = 3 - 1

Simplify

4 cos (2 t) = 2

4 cos (2 t) = 2

Divide both sides by

4

4 cos (2 t) = 2

Divide both sides by

4

\frac{4 cos ( 2 t )}{4} = \frac{2}{4}

Simplify

cos (2 t) = \frac{1}{2}

cos (2 t) = \frac{1}{2}

General solutions for

cos (2 t) = \frac{1}{2}

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}

2 t = \frac{π}{3} + 2 πn, 2 t = \frac{5 π}{3} + 2 πn

2 t = \frac{π}{3} + 2 πn, 2 t = \frac{5 π}{3} + 2 πn

Solve

2 t = \frac{π}{3} + 2 πn : t = \frac{π}{6} + πn

2 t = \frac{π}{3} + 2 πn

Divide both sides by

2

2 t = \frac{π}{3} + 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 t}{2} = \frac{\frac{π}{3}}{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}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

t = \frac{π}{6} + πn

t = \frac{π}{6} + πn

t = \frac{π}{6} + πn

Solve

2 t = \frac{5 π}{3} + 2 πn : t = \frac{5 π}{6} + πn

2 t = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 t = \frac{5 π}{3} + 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 t}{2} = \frac{\frac{5 π}{3}}{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{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

t = \frac{5 π}{6} + πn

t = \frac{5 π}{6} + πn

t = \frac{5 π}{6} + πn

t = \frac{π}{6} + πn, t = \frac{5 π}{6} + πn

Solutions for the range

0

No Solution for t \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos(2t)+1=3,0, pi/2 ?
The general solution for 4cos(2t)+1=3,0, pi/2 is No Solution for t\in\mathbb{R}