What is the general solution for 2cos(2c)=((0)-(0))/((pi)-(-pi)) ?

The general solution for 2cos(2c)=((0)-(0))/((pi)-(-pi)) is c= pi/4+pin,c=(3pi)/4+pin

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2cos(2c)=((0)-(0))/((pi)-(-pi))

Solution

2 cos (2 c) = \frac{( 0 ) - ( 0 )}{( π ) - ( - π )}

Solution

c = \frac{π}{4} + πn, c = \frac{3 π}{4} + πn

Degrees

c = 4 5^{\circ} + 18 0^{\circ} n, c = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (2 c) = \frac{( 0 ) - ( 0 )}{( π ) - ( - π )}

Apply rule

- (- a) = a

2 cos (2 c) = \frac{0 - 0}{π + π}

Subtract the numbers:

0 - 0 = 0

2 cos (2 c) = \frac{0}{π + π}

Add similar elements:

π + π = 2 π

2 cos (2 c) = \frac{0}{2 π}

Apply rule

\frac{0}{a} = 0, a \neq = 0

2 cos (2 c) = 0

Divide both sides by

2

2 cos (2 c) = 0

Divide both sides by

2

\frac{2 cos ( 2 c )}{2} = \frac{0}{2}

Simplify

cos (2 c) = 0

cos (2 c) = 0

General solutions for

cos (2 c) = 0

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

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

Solve

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

2 c = \frac{π}{2} + 2 πn

Divide both sides by

2

2 c = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 c}{2} : c

\frac{2 c}{2}

Divide the numbers:

\frac{2}{2} = 1

= c

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

c = \frac{π}{4} + πn

c = \frac{π}{4} + πn

c = \frac{π}{4} + πn

Solve

2 c = \frac{3 π}{2} + 2 πn : c = \frac{3 π}{4} + πn

2 c = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 c = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 c}{2} : c

\frac{2 c}{2}

Divide the numbers:

\frac{2}{2} = 1

= c

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

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

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

c = \frac{3 π}{4} + πn

c = \frac{3 π}{4} + πn

c = \frac{3 π}{4} + πn

c = \frac{π}{4} + πn, c = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2c)=((0)-(0))/((pi)-(-pi)) ?
The general solution for 2cos(2c)=((0)-(0))/((pi)-(-pi)) is c= pi/4+pin,c=(3pi)/4+pin