What is the general solution for 4cos(2x-pi)-1=1 ?

The general solution for 4cos(2x-pi)-1=1 is x=pin+pi/2+pi/6 ,x=pin+pi/2+(5pi)/6

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4cos(2x-pi)-1=1

Solution

4 cos (2 x - π) - 1 = 1

Solution

x = πn + \frac{π}{2} + \frac{π}{6}, x = πn + \frac{π}{2} + \frac{5 π}{6}

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n, x = 24 0^{\circ} + 18 0^{\circ} n

Solution steps

4 cos (2 x - π) - 1 = 1

Move

1

to the right side

4 cos (2 x - π) - 1 = 1

Add

1

to both sides

4 cos (2 x - π) - 1 + 1 = 1 + 1

Simplify

4 cos (2 x - π) = 2

4 cos (2 x - π) = 2

Divide both sides by

4

4 cos (2 x - π) = 2

Divide both sides by

4

\frac{4 cos ( 2 x - π )}{4} = \frac{2}{4}

Simplify

cos (2 x - π) = \frac{1}{2}

cos (2 x - π) = \frac{1}{2}

General solutions for

cos (2 x - π) = \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 x - π = \frac{π}{3} + 2 πn, 2 x - π = \frac{5 π}{3} + 2 πn

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

Solve

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

2 x - π = \frac{π}{3} + 2 πn

Move

π

to the right side

2 x - π = \frac{π}{3} + 2 πn

Add

π

to both sides

2 x - π + π = \frac{π}{3} + 2 πn + π

Simplify

2 x = \frac{π}{3} + 2 πn + π

2 x = \frac{π}{3} + 2 πn + π

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn + π

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

Group like terms

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\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{π}{2} + πn + \frac{π}{6}

Group like terms

= πn + \frac{π}{2} + \frac{π}{6}

x = πn + \frac{π}{2} + \frac{π}{6}

x = πn + \frac{π}{2} + \frac{π}{6}

x = πn + \frac{π}{2} + \frac{π}{6}

Solve

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

2 x - π = \frac{5 π}{3} + 2 πn

Move

π

to the right side

2 x - π = \frac{5 π}{3} + 2 πn

Add

π

to both sides

2 x - π + π = \frac{5 π}{3} + 2 πn + π

Simplify

2 x = \frac{5 π}{3} + 2 πn + π

2 x = \frac{5 π}{3} + 2 πn + π

Divide both sides by

2

2 x = \frac{5 π}{3} + 2 πn + π

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

Group like terms

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\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{π}{2} + πn + \frac{5 π}{6}

Group like terms

= πn + \frac{π}{2} + \frac{5 π}{6}

x = πn + \frac{π}{2} + \frac{5 π}{6}

x = πn + \frac{π}{2} + \frac{5 π}{6}

x = πn + \frac{π}{2} + \frac{5 π}{6}

x = πn + \frac{π}{2} + \frac{π}{6}, x = πn + \frac{π}{2} + \frac{5 π}{6}

Popular Examples

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csc (x) = 4 sin (x)

tan(3+x)=0

tan (3 + x) = 0

2cot^2(x)-9csc(x)=3

2 cot^{2} (x) - 9 csc (x) = 3

1=sin(2θ)

1 = sin (2 θ)

csc(X)-cot(X)=5

csc (X) - cot (X) = 5

Frequently Asked Questions (FAQ)

What is the general solution for 4cos(2x-pi)-1=1 ?
The general solution for 4cos(2x-pi)-1=1 is x=pin+pi/2+pi/6 ,x=pin+pi/2+(5pi)/6