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Popular >

sqrt(3)tan(3θ)-1=0,0<= θ<= 2pi

Solution

3 tan (3 θ) - 1 = 0, 0 \leq θ \leq 2 π

Solution

θ = \frac{π}{18}, θ = \frac{7 π}{18}, θ = \frac{13 π}{18}, θ = \frac{19 π}{18}, θ = \frac{25 π}{18}

+1

Degrees

θ = 1 0^{\circ}, θ = 7 0^{\circ}, θ = 13 0^{\circ}, θ = 19 0^{\circ}, θ = 25 0^{\circ}

Solution steps

3 tan (3 θ) - 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

3 tan (3 θ) - 1 = 0

Add

1

to both sides

3 tan (3 θ) - 1 + 1 = 0 + 1

Simplify

3 tan (3 θ) = 1

3 tan (3 θ) = 1

Divide both sides by

3

3 tan (3 θ) = 1

Divide both sides by

3

\frac{3 tan ( 3 θ )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( 3 θ )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( 3 θ )}{3} : tan (3 θ)

\frac{3 tan ( 3 θ )}{3}

Cancel the common factor:

3

= tan (3 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 (3 θ) = \frac{3}{3}

tan (3 θ) = \frac{3}{3}

tan (3 θ) = \frac{3}{3}

General solutions for

tan (3 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}

3 θ = \frac{π}{6} + πn

3 θ = \frac{π}{6} + πn

Solve

3 θ = \frac{π}{6} + πn : θ = \frac{π}{18} + \frac{πn}{3}

3 θ = \frac{π}{6} + πn

Divide both sides by

3

3 θ = \frac{π}{6} + πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{π}{6}}{3} + \frac{πn}{3} : \frac{π}{18} + \frac{πn}{3}

\frac{\frac{π}{6}}{3} + \frac{πn}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= \frac{π}{18} + \frac{πn}{3}

θ = \frac{π}{18} + \frac{πn}{3}

θ = \frac{π}{18} + \frac{πn}{3}

θ = \frac{π}{18} + \frac{πn}{3}

θ = \frac{π}{18} + \frac{πn}{3}

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{18}, θ = \frac{7 π}{18}, θ = \frac{13 π}{18}, θ = \frac{19 π}{18}, θ = \frac{25 π}{18}

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