What is the general solution for sqrt(3)tan(x/2)-1=0 ?

The general solution for sqrt(3)tan(x/2)-1=0 is x= pi/3+2pin

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

Solution

3 tan (\frac{x}{2}) - 1 = 0

Solution

x = \frac{π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n

Solution steps

3 tan (\frac{x}{2}) - 1 = 0

Move

1

to the right side

3 tan (\frac{x}{2}) - 1 = 0

Add

1

to both sides

3 tan (\frac{x}{2}) - 1 + 1 = 0 + 1

Simplify

3 tan (\frac{x}{2}) = 1

3 tan (\frac{x}{2}) = 1

Divide both sides by

3

3 tan (\frac{x}{2}) = 1

Divide both sides by

3

\frac{3 tan ( \frac{x}{2} )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( \frac{x}{2} )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( \frac{x}{2} )}{3} : tan (\frac{x}{2})

\frac{3 tan ( \frac{x}{2} )}{3}

Cancel the common factor:

3

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

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

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

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

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

General solutions for

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

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

\frac{x}{2} = \frac{π}{6} + πn

\frac{x}{2} = \frac{π}{6} + πn

Solve

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

\frac{x}{2} = \frac{π}{6} + πn

Multiply both sides by

2

\frac{x}{2} = \frac{π}{6} + πn

Multiply both sides by

2

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

Simplify

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

2 \cdot \frac{π}{6} + 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}

= \frac{π}{3} + 2 πn

x = \frac{π}{3} + 2 πn

x = \frac{π}{3} + 2 πn

x = \frac{π}{3} + 2 πn

x = \frac{π}{3} + 2 πn

Popular Examples

cos(2x)=0.5

cos (2 x) = 0.5

cos^2(x)=-1

cos^{2} (x) = - 1

cos(2x)=1-sin(x)

cos (2 x) = 1 - sin (x)

-cos(x)=-sin^2(x)-1

- cos (x) = - sin^{2} (x) - 1

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

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

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)tan(x/2)-1=0 ?
The general solution for sqrt(3)tan(x/2)-1=0 is x= pi/3+2pin