What is the general solution for -8cos(2x)=0,(-pi,pi) ?

The general solution for -8cos(2x)=0,(-pi,pi) is x= pi/4 ,x=(3pi)/4 ,x=-(3pi)/4 ,x=-pi/4

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-8cos(2x)=0,(-pi,pi)

Solution

- 8 cos (2 x) = 0, (- π, π)

Solution

x = \frac{π}{4}, x = \frac{3 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}

Degrees

x = 4 5^{\circ}, x = 13 5^{\circ}, x = - 13 5^{\circ}, x = - 4 5^{\circ}

Solution steps

- 8 cos (2 x) = 0, (- π, π)

Divide both sides by

- 8

- 8 cos (2 x) = 0

Divide both sides by

- 8

\frac{- 8 cos ( 2 x )}{- 8} = \frac{0}{- 8}

Simplify

cos (2 x) = 0

cos (2 x) = 0

General solutions for

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

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

Solve

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

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Solve

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

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

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

Solutions for the range

(- π, π)

x = \frac{π}{4}, x = \frac{3 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}

Popular Examples

1.66=sin(x)

1.66 = sin (x)

sin(x)=(3pi)/2

sin (x) = \frac{3 π}{2}

2sec^2(x)=3-tan(x),0<= x<= 2pi

2 sec^{2} (x) = 3 - tan (x), 0 \leq x \leq 2 π

1+cos(2θ)=-3cos(θ)

1 + cos (2 θ) = - 3 cos (θ)

tan(θ)= 6/0

tan (θ) = \frac{6}{0}

Frequently Asked Questions (FAQ)

What is the general solution for -8cos(2x)=0,(-pi,pi) ?
The general solution for -8cos(2x)=0,(-pi,pi) is x= pi/4 ,x=(3pi)/4 ,x=-(3pi)/4 ,x=-pi/4