What is the general solution for 2cos(x/2)=sqrt(3),0<= x<= 2pi ?

Q: What is the general solution for 2cos(x/2)=sqrt(3),0<= x<= 2pi ?

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

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2cos(x/2)=sqrt(3),0<= x<= 2pi

Solution

2 cos (\frac{x}{2}) = 3, 0 \leq x \leq 2 π

Solution

x = \frac{π}{3}

Degrees

x = 6 0^{\circ}

Solution steps

2 cos (\frac{x}{2}) = 3, 0 \leq x \leq 2 π

Divide both sides by

2

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

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

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

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

Solve

\frac{x}{2} = \frac{π}{6} + 2 πn : x = \frac{π}{3} + 4 πn

\frac{x}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{π}{6} + 2 \cdot 2 πn : \frac{π}{3} + 4 πn

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

x = \frac{π}{3} + 4 πn

Solve

\frac{x}{2} = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{3} + 4 πn

\frac{x}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

\frac{x}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

Multiply fractions:

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

= \frac{11 π 2}{6}

Multiply the numbers:

11 \cdot 2 = 22

= \frac{22 π}{6}

Cancel the common factor:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{11 π}{3} + 4 πn

x = \frac{π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}

Popular Examples

sin(A)= 4/5

sin (A) = \frac{4}{5}

3tan^2(x)-1=0,0<= x<= 2pi

3 tan^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

sin(2x)=4cos(x)

sin (2 x) = 4 cos (x)

sqrt(3)tan(2x)-1=0

3 tan (2 x) - 1 = 0

cos(2x+pi)=0

cos (2 x + π) = 0

Frequently Asked Questions (FAQ)

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