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))

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

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

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

cos (2 c) = 0

General solutions for

cos (2 c) = 0

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

Solve

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

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

- 5 cos (2 θ) - 9 sin (θ) + 8 = - sin (θ)

3 + 3 cos (θ) = 3 + 3 sin (θ)

sinh (x) = \frac{4}{3}

x, f^{- 1} = arctan (x^{3} - 1)

3 cos (B) + 3 = 0

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