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

The general solution for sqrt(3)tan(θ)-1=0,0<= θ<= 2pi is θ= pi/6 ,θ=(7pi)/6

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sqrt(3)tan(θ)-1=0,0<= θ<= 2pi

3 tan (θ) - 1 = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{7 π}{6}

Degrees

θ = 3 0^{\circ}, θ = 21 0^{\circ}

Solution steps

3 tan (θ) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

3 tan (θ) - 1 = 0

Add

1

to both sides

3 tan (θ) - 1 + 1 = 0 + 1

Simplify

3 tan (θ) = 1

3 tan (θ) = 1

Divide both sides by

3

3 tan (θ) = 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( θ )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

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 (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \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}

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{7 π}{6}

- cos (\frac{π}{2})

sin^{3} (\frac{π}{2})

sin (3 θ) = 1

4 sin^{2} (x) = 3

sin (\frac{7 π}{2})

What is the general solution for sqrt(3)tan(θ)-1=0,0<= θ<= 2pi ?
The general solution for sqrt(3)tan(θ)-1=0,0<= θ<= 2pi is θ= pi/6 ,θ=(7pi)/6