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

Q: What is the general solution for cos(2x)=0,0<= x<= 2pi ?

The general solution for cos(2x)=0,0<= x<= 2pi is x= pi/4 ,x=(3pi)/4 ,x=(5pi)/4 ,x=(7pi)/4

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cos(2x)=0,0<= x<= 2pi

Solution

cos (2 x) = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Degrees

x = 4 5^{\circ}, x = 13 5^{\circ}, x = 22 5^{\circ}, x = 31 5^{\circ}

Solution steps

cos (2 x) = 0, 0 \leq x \leq 2 π

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

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Popular Examples

cot^2(θ)-csc(θ)=1

cot^{2} (θ) - csc (θ) = 1

cos(x/2)=-(sqrt(3))/2

cos (\frac{x}{2}) = - \frac{3}{2}

tan(x)=0.2

tan (x) = 0.2

sin(2t)+sin(t)=0

sin (2 t) + sin (t) = 0

2sin^2(x)=3(cos(x)+1)

2 sin^{2} (x) = 3 (cos (x) + 1)

Frequently Asked Questions (FAQ)

What is the general solution for cos(2x)=0,0<= x<= 2pi ?
The general solution for cos(2x)=0,0<= x<= 2pi is x= pi/4 ,x=(3pi)/4 ,x=(5pi)/4 ,x=(7pi)/4