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Popular >

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

Solution

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ \leq 2 π

Solution

θ = 0, θ = π, θ = 2 π, θ = \frac{π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 0^{\circ}, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 3 0^{\circ}, θ = 33 0^{\circ}

Solution steps

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ \leq 2 π

Subtract

\frac{2 3}{3} sin (θ)

from both sides

tan (θ) - \frac{2 sin ( θ )}{3} = 0

Simplify

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

tan (θ) - \frac{2 sin ( θ )}{3}

Convert element to fraction:

tan (θ) = \frac{tan ( θ ) 3}{3}

= \frac{tan ( θ ) 3}{3} - \frac{2 sin ( θ )}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( θ ) 3 - 2 sin ( θ )}{3}

\frac{3 tan ( θ ) - 2 sin ( θ )}{3} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 tan (θ) - 2 sin (θ) = 0

Express with sin, cos

- 2 sin (θ) + 3 tan (θ)

Use the basic trigonometric identity:

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

= - 2 sin (θ) + 3 \frac{sin ( θ )}{cos ( θ )}

Simplify

- 2 sin (θ) + 3 \frac{sin ( θ )}{cos ( θ )} : \frac{- 2 sin ( θ ) cos ( θ ) + 3 sin ( θ )}{cos ( θ )}

- 2 sin (θ) + 3 \frac{sin ( θ )}{cos ( θ )}

Multiply

3 \frac{sin ( θ )}{cos ( θ )} : \frac{3 sin ( θ )}{cos ( θ )}

3 \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) 3}{cos ( θ )}

= - 2 sin (θ) + \frac{3 sin ( θ )}{cos ( θ )}

Convert element to fraction:

2 sin (θ) = \frac{2 sin ( θ ) cos ( θ )}{cos ( θ )}

= - \frac{2 sin ( θ ) cos ( θ )}{cos ( θ )} + \frac{sin ( θ ) 3}{cos ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 2 sin ( θ ) cos ( θ ) + sin ( θ ) 3}{cos ( θ )}

= \frac{- 2 sin ( θ ) cos ( θ ) + 3 sin ( θ )}{cos ( θ )}

\frac{sin ( θ ) 3 - 2 cos ( θ ) sin ( θ )}{cos ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (θ) 3 - 2 cos (θ) sin (θ) = 0

Factor

sin (θ) 3 - 2 cos (θ) sin (θ) : sin (θ) (3 - 2 cos (θ))

sin (θ) 3 - 2 cos (θ) sin (θ)

Factor out common term

sin (θ)

= sin (θ) (3 - 2 cos (θ))

sin (θ) (3 - 2 cos (θ)) = 0

Solving each part separately

sin (θ) = 0 or 3 - 2 cos (θ) = 0

sin (θ) = 0, 0 \leq θ \leq 2 π : θ = 0, θ = π, θ = 2 π

sin (θ) = 0, 0 \leq θ \leq 2 π

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π

3 - 2 cos (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{11 π}{6}

3 - 2 cos (θ) = 0, 0 \leq θ \leq 2 π

Move

3

to the right side

3 - 2 cos (θ) = 0

Subtract

3

from both sides

3 - 2 cos (θ) - 3 = 0 - 3

Simplify

- 2 cos (θ) = - 3

- 2 cos (θ) = - 3

Divide both sides by

- 2

- 2 cos (θ) = - 3

Divide both sides by

- 2

\frac{- 2 cos ( θ )}{- 2} = \frac{- 3}{- 2}

Simplify

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

General solutions for

cos (θ) = \frac{3}{2}

cos (x)

periodicity table with

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = 0, θ = π, θ = 2 π, θ = \frac{π}{6}, θ = \frac{11 π}{6}

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