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

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

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Popular >

sqrt(3)tan(θ/2)-1=0

Solution

3 tan (\frac{θ}{2}) - 1 = 0

Solution

θ = \frac{π}{3} + 2 πn

Degrees

θ = 6 0^{\circ} + 36 0^{\circ} n

Solution steps

3 tan (\frac{θ}{2}) - 1 = 0

Move

1

to the right side

3 tan (\frac{θ}{2}) - 1 = 0

Add

1

to both sides

3 tan (\frac{θ}{2}) - 1 + 1 = 0 + 1

Simplify

3 tan (\frac{θ}{2}) = 1

3 tan (\frac{θ}{2}) = 1

Divide both sides by

3

3 tan (\frac{θ}{2}) = 1

Divide both sides by

3

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

Simplify

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

Simplify

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

\frac{3 tan ( \frac{θ}{2} )}{3}

Cancel the common factor:

3

= tan (\frac{θ}{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{θ}{2}) = \frac{3}{3}

tan (\frac{θ}{2}) = \frac{3}{3}

tan (\frac{θ}{2}) = \frac{3}{3}

General solutions for

tan (\frac{θ}{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{θ}{2} = \frac{π}{6} + πn

\frac{θ}{2} = \frac{π}{6} + πn

Solve

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

\frac{θ}{2} = \frac{π}{6} + πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{6} + πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

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

θ = \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn

Popular Examples

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

tan (\frac{x}{2}) = 3, 0 \leq x < 2 π

sin(x)= 4/(cos(x))

sin (x) = \frac{4}{cos ( x )}

tan^3(x)-2tan(x)=tan(x)

tan^{3} (x) - 2 tan (x) = tan (x)

sin^2(x)+2cos^2(x)=0

sin^{2} (x) + 2 cos^{2} (x) = 0

sin(x+16)=cos(11x+2)

sin (x + 1 6^{\circ}) = cos (11 x + 2)

Frequently Asked Questions (FAQ)

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