What is the general solution for sqrt(3)tan(θ)=1 ?

The general solution for sqrt(3)tan(θ)=1 is θ= pi/6+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(3)tan(θ)=1

3 tan (θ) = 1

θ = \frac{π}{6} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

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

2 cos (3 x) - 1 = 0, 0 \leq x \leq 2 π

16 cos^{2} (θ) = 4

sin (3 θ + 72) = cos (4 8^{\circ})

tan (x) = \frac{1}{5}

sec (x) = 2 cos (x)

What is the general solution for sqrt(3)tan(θ)=1 ?
The general solution for sqrt(3)tan(θ)=1 is θ= pi/6+pin