What is the general solution for 2cos(2x)-sqrt(3)=0 ?

The general solution for 2cos(2x)-sqrt(3)=0 is x= pi/(12)+pin,x=(11pi)/(12)+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(2x)-sqrt(3)=0

Solution

2 cos (2 x) - 3 = 0

Solution

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

Degrees

x = 1 5^{\circ} + 18 0^{\circ} n, x = 16 5^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (2 x) - 3 = 0

Move

3

to the right side

2 cos (2 x) - 3 = 0

Add

3

to both sides

2 cos (2 x) - 3 + 3 = 0 + 3

Simplify

2 cos (2 x) = 3

2 cos (2 x) = 3

Divide both sides by

2

2 cos (2 x) = 3

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (2 x) = \frac{3}{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 = \frac{π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

2 x = \frac{π}{6} + 2 πn, 2 x = \frac{11 π}{6} + 2 πn

Solve

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{π}{6} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{π}{6}}{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}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

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

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

Solve

2 x = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{12} + πn

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{11 π}{6} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{11 π}{6}}{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{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{11 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{11 π}{12} + πn

Popular Examples

sin(x)-cos(x)=0

sin (x) - cos (x) = 0

cos(2x)-cos(x)=0

cos (2 x) - cos (x) = 0

csc(-pi/2)

csc (- \frac{π}{2})

tan^2(x)=1

tan^{2} (x) = 1

sec((11pi)/6)

sec (\frac{11 π}{6})

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)-sqrt(3)=0 ?
The general solution for 2cos(2x)-sqrt(3)=0 is x= pi/(12)+pin,x=(11pi)/(12)+pin