What is the general solution for tan(θ+20)tan(90-3θ)=1 ?

The general solution for tan(θ+20)tan(90-3θ)=1 is θ=-180n+10,θ=-80-180n

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Popular Trigonometry >

tan(θ+20)tan(90-3θ)=1

Solution

tan (θ + 2 0^{\circ}) tan (9 0^{\circ} - 3 θ) = 1

Solution

θ = - 18 0^{\circ} n + 1 0^{\circ}, θ = - 8 0^{\circ} - 18 0^{\circ} n

Radians

θ = \frac{π}{18} - πn, θ = - \frac{4 π}{9} - πn

Solution steps

tan (θ + 2 0^{\circ}) tan (9 0^{\circ} - 3 θ) = 1

Rewrite using trig identities

tan (θ + 2 0^{\circ}) tan (9 0^{\circ} - 3 θ) = 1

Rewrite using trig identities

tan (9 0^{\circ} - 3 θ)

Use the basic trigonometric identity:

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

= \frac{sin ( 9 0 ^{\circ} - 3 θ )}{cos ( 9 0 ^{\circ} - 3 θ )}

Use the Angle Difference identity:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( 9 0 ^{\circ} ) cos ( 3 θ ) - cos ( 9 0 ^{\circ} ) sin ( 3 θ )}{cos ( 9 0 ^{\circ} - 3 θ )}

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( 9 0 ^{\circ} ) cos ( 3 θ ) - cos ( 9 0 ^{\circ} ) sin ( 3 θ )}{cos ( 9 0 ^{\circ} ) cos ( 3 θ ) + sin ( 9 0 ^{\circ} ) sin ( 3 θ )}

Simplify

\frac{sin ( 9 0 ^{\circ} ) cos ( 3 θ ) - cos ( 9 0 ^{\circ} ) sin ( 3 θ )}{cos ( 9 0 ^{\circ} ) cos ( 3 θ ) + sin ( 9 0 ^{\circ} ) sin ( 3 θ )} : \frac{cos ( 3 θ )}{sin ( 3 θ )}

\frac{sin ( 9 0 ^{\circ} ) cos ( 3 θ ) - cos ( 9 0 ^{\circ} ) sin ( 3 θ )}{cos ( 9 0 ^{\circ} ) cos ( 3 θ ) + sin ( 9 0 ^{\circ} ) sin ( 3 θ )}

sin (9 0^{\circ}) cos (3 θ) - cos (9 0^{\circ}) sin (3 θ) = cos (3 θ)

sin (9 0^{\circ}) cos (3 θ) - cos (9 0^{\circ}) sin (3 θ)

sin (9 0^{\circ}) cos (3 θ) = cos (3 θ)

sin (9 0^{\circ}) cos (3 θ)

Simplify

sin (9 0^{\circ}) : 1

sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1 \cdot cos (3 θ)

Multiply:

1 \cdot cos (3 θ) = cos (3 θ)

= cos (3 θ)

cos (9 0^{\circ}) sin (3 θ) = 0

cos (9 0^{\circ}) sin (3 θ)

Simplify

cos (9 0^{\circ}) : 0

cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0 \cdot sin (3 θ)

Apply rule

0 \cdot a = 0

= 0

= cos (3 θ) - 0

cos (3 θ) - 0 = cos (3 θ)

= cos (3 θ)

= \frac{cos ( 3 θ )}{cos ( 9 0 ^{\circ} ) cos ( 3 θ ) + sin ( 9 0 ^{\circ} ) sin ( 3 θ )}

cos (9 0^{\circ}) cos (3 θ) + sin (9 0^{\circ}) sin (3 θ) = sin (3 θ)

cos (9 0^{\circ}) cos (3 θ) + sin (9 0^{\circ}) sin (3 θ)

cos (9 0^{\circ}) cos (3 θ) = 0

cos (9 0^{\circ}) cos (3 θ)

Simplify

cos (9 0^{\circ}) : 0

cos (9 0^{\circ})

Use the following trivial identity:

cos (9 0^{\circ}) = 0

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0 \cdot cos (3 θ)

Apply rule

0 \cdot a = 0

= 0

sin (9 0^{\circ}) sin (3 θ) = sin (3 θ)

sin (9 0^{\circ}) sin (3 θ)

Simplify

sin (9 0^{\circ}) : 1

sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1 \cdot sin (3 θ)

Multiply:

1 \cdot sin (3 θ) = sin (3 θ)

= sin (3 θ)

= 0 + sin (3 θ)

0 + sin (3 θ) = sin (3 θ)

= sin (3 θ)

= \frac{cos ( 3 θ )}{sin ( 3 θ )}

= \frac{cos ( 3 θ )}{sin ( 3 θ )}

tan (θ + 2 0^{\circ}) \frac{cos ( 3 θ )}{sin ( 3 θ )} = 1

Simplify

tan (θ + 2 0^{\circ}) \frac{cos ( 3 θ )}{sin ( 3 θ )} : \frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )}

tan (θ + 2 0^{\circ}) \frac{cos ( 3 θ )}{sin ( 3 θ )}

Multiply fractions:

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

= \frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )}

\frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )} = 1

\frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )} = 1

Subtract

1

from both sides

\frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )} - 1 = 0

Simplify

\frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )} - 1 : \frac{cos ( 3 θ ) tan ( \frac{9 θ + 18 0 ^{\circ}}{9} ) - sin ( 3 θ )}{sin ( 3 θ )}

\frac{cos ( 3 θ ) tan ( θ + 2 0 ^{\circ} )}{sin ( 3 θ )} - 1

Join

θ + 2 0^{\circ} : \frac{9 θ + 18 0 ^{\circ}}{9}

θ + 2 0^{\circ}

Convert element to fraction:

θ = \frac{θ 9}{9}

= \frac{θ \cdot 9}{9} + 2 0^{\circ}

Since the denominators are equal, combine the fractions:

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

= \frac{θ \cdot 9 + 18 0 ^{\circ}}{9}

= \frac{cos ( 3 θ ) tan ( \frac{9 θ + 18 0 ^{\circ}}{9} )}{sin ( 3 θ )} - 1

Convert element to fraction:

1 = \frac{1 sin ( 3 θ )}{sin ( 3 θ )}

= \frac{cos ( 3 θ ) tan ( \frac{θ \cdot 9 + 18 0 ^{\circ}}{9} )}{sin ( 3 θ )} - \frac{1 \cdot sin ( 3 θ )}{sin ( 3 θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( 3 θ ) tan ( \frac{θ \cdot 9 + 18 0 ^{\circ}}{9} ) - 1 \cdot sin ( 3 θ )}{sin ( 3 θ )}

Multiply:

1 \cdot sin (3 θ) = sin (3 θ)

= \frac{cos ( 3 θ ) tan ( \frac{9 θ + 18 0 ^{\circ}}{9} ) - sin ( 3 θ )}{sin ( 3 θ )}

\frac{cos ( 3 θ ) tan ( \frac{9 θ + 18 0 ^{\circ}}{9} ) - sin ( 3 θ )}{sin ( 3 θ )} = 0

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

cos (3 θ) tan (\frac{9 θ + 18 0 ^{\circ}}{9}) - sin (3 θ) = 0

Express with sin, cos

- sin (3 θ) + cos (3 θ) tan (\frac{18 0 ^{\circ} + 9 θ}{9})

Use the basic trigonometric identity:

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

= - sin (3 θ) + cos (3 θ) \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

Simplify

- sin (3 θ) + cos (3 θ) \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )} : \frac{- sin ( 3 θ ) cos ( \frac{18 0 ^{\circ} + 9 θ}{9} ) + sin ( \frac{18 0 ^{\circ} + 9 θ}{9} ) cos ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

- sin (3 θ) + cos (3 θ) \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

Multiply

cos (3 θ) \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )} : \frac{sin ( \frac{9 θ + 18 0 ^{\circ}}{9} ) cos ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

cos (3 θ) \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

Multiply fractions:

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

= \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} ) cos ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

= - sin (3 θ) + \frac{sin ( \frac{9 θ + 18 0 ^{\circ}}{9} ) cos ( 3 θ )}{cos ( \frac{9 θ + 18 0 ^{\circ}}{9} )}

Convert element to fraction:

sin (3 θ) = \frac{sin ( 3 θ ) cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

= - \frac{sin ( 3 θ ) cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )} + \frac{sin ( \frac{18 0 ^{\circ} + 9 θ}{9} ) cos ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

Since the denominators are equal, combine the fractions:

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

= \frac{- sin ( 3 θ ) cos ( \frac{18 0 ^{\circ} + 9 θ}{9} ) + sin ( \frac{18 0 ^{\circ} + 9 θ}{9} ) cos ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

= \frac{- sin ( 3 θ ) cos ( \frac{18 0 ^{\circ} + 9 θ}{9} ) + sin ( \frac{18 0 ^{\circ} + 9 θ}{9} ) cos ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )}

\frac{cos ( 3 θ ) sin ( \frac{18 0 ^{\circ} + 9 θ}{9} ) - cos ( \frac{18 0 ^{\circ} + 9 θ}{9} ) sin ( 3 θ )}{cos ( \frac{18 0 ^{\circ} + 9 θ}{9} )} = 0

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

cos (3 θ) sin (\frac{18 0 ^{\circ} + 9 θ}{9}) - cos (\frac{18 0 ^{\circ} + 9 θ}{9}) sin (3 θ) = 0

Rewrite using trig identities

cos (3 θ) sin (\frac{18 0 ^{\circ} + 9 θ}{9}) - cos (\frac{18 0 ^{\circ} + 9 θ}{9}) sin (3 θ)

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ)

sin (\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ) = 0

General solutions for

sin (\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ) = 0

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 0 + 36 0^{\circ} n, \frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 18 0^{\circ} + 36 0^{\circ} n

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 0 + 36 0^{\circ} n, \frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 18 0^{\circ} + 36 0^{\circ} n

Solve

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 0 + 36 0^{\circ} n : θ = - 18 0^{\circ} n + 1 0^{\circ}

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 0 + 36 0^{\circ} n

0 + 36 0^{\circ} n = 36 0^{\circ} n

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 36 0^{\circ} n

Multiply both sides by

9

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 36 0^{\circ} n

Multiply both sides by

9

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9 - 3 θ \cdot 9 = 36 0^{\circ} n \cdot 9

Simplify

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9 - 3 θ \cdot 9 = 36 0^{\circ} n \cdot 9

Simplify

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9 : 18 0^{\circ} + 9 θ

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9

Multiply fractions:

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

= \frac{( 18 0 ^{\circ} + 9 θ ) \cdot 9}{9}

Cancel the common factor:

9

= 18 0^{\circ} + 9 θ

Simplify

- 3 θ \cdot 9 : - 27 θ

- 3 θ \cdot 9

Multiply the numbers:

3 \cdot 9 = 27

= - 27 θ

Simplify

36 0^{\circ} n \cdot 9 : 324 0^{\circ} n

36 0^{\circ} n \cdot 9

Multiply the numbers:

2 \cdot 9 = 18

= 324 0^{\circ} n

18 0^{\circ} + 9 θ - 27 θ = 324 0^{\circ} n

18 0^{\circ} - 18 θ = 324 0^{\circ} n

18 0^{\circ} - 18 θ = 324 0^{\circ} n

18 0^{\circ} - 18 θ = 324 0^{\circ} n

Move

18 0^{\circ}

to the right side

18 0^{\circ} - 18 θ = 324 0^{\circ} n

Subtract

18 0^{\circ}

from both sides

18 0^{\circ} - 18 θ - 18 0^{\circ} = 324 0^{\circ} n - 18 0^{\circ}

Simplify

- 18 θ = 324 0^{\circ} n - 18 0^{\circ}

- 18 θ = 324 0^{\circ} n - 18 0^{\circ}

Divide both sides by

- 18

- 18 θ = 324 0^{\circ} n - 18 0^{\circ}

Divide both sides by

- 18

\frac{- 18 θ}{- 18} = \frac{324 0 ^{\circ} n}{- 18} - \frac{18 0 ^{\circ}}{- 18}

Simplify

\frac{- 18 θ}{- 18} = \frac{324 0 ^{\circ} n}{- 18} - \frac{18 0 ^{\circ}}{- 18}

Simplify

\frac{- 18 θ}{- 18} : θ

\frac{- 18 θ}{- 18}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{18 θ}{18}

Divide the numbers:

\frac{18}{18} = 1

= θ

Simplify

\frac{324 0 ^{\circ} n}{- 18} - \frac{18 0 ^{\circ}}{- 18} : - 18 0^{\circ} n + 1 0^{\circ}

\frac{324 0 ^{\circ} n}{- 18} - \frac{18 0 ^{\circ}}{- 18}

\frac{324 0 ^{\circ} n}{- 18} = - 18 0^{\circ} n

\frac{324 0 ^{\circ} n}{- 18}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{324 0 ^{\circ} n}{18}

Divide the numbers:

\frac{18}{18} = 1

= - 18 0^{\circ} n

= - 18 0^{\circ} n - \frac{18 0 ^{\circ}}{- 18}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - 18 0^{\circ} n - (- 1 0^{\circ})

Apply rule

- (- a) = a

= - 18 0^{\circ} n + 1 0^{\circ}

θ = - 18 0^{\circ} n + 1 0^{\circ}

θ = - 18 0^{\circ} n + 1 0^{\circ}

θ = - 18 0^{\circ} n + 1 0^{\circ}

Solve

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 18 0^{\circ} + 36 0^{\circ} n : θ = - 8 0^{\circ} - 18 0^{\circ} n

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 18 0^{\circ} + 36 0^{\circ} n

Multiply both sides by

9

\frac{18 0 ^{\circ} + 9 θ}{9} - 3 θ = 18 0^{\circ} + 36 0^{\circ} n

Multiply both sides by

9

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9 - 3 θ \cdot 9 = 18 0^{\circ} 9 + 36 0^{\circ} n \cdot 9

Simplify

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9 - 3 θ \cdot 9 = 18 0^{\circ} 9 + 36 0^{\circ} n \cdot 9

Simplify

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9 : 18 0^{\circ} + 9 θ

\frac{18 0 ^{\circ} + 9 θ}{9} \cdot 9

Multiply fractions:

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

= \frac{( 18 0 ^{\circ} + 9 θ ) \cdot 9}{9}

Cancel the common factor:

9

= 18 0^{\circ} + 9 θ

Simplify

- 3 θ \cdot 9 : - 27 θ

- 3 θ \cdot 9

Multiply the numbers:

3 \cdot 9 = 27

= - 27 θ

Simplify

18 0^{\circ} 9 : 162 0^{\circ}

18 0^{\circ} 9

Apply the commutative law:

18 0^{\circ} 9 = 162 0^{\circ}

162 0^{\circ}

Simplify

36 0^{\circ} n \cdot 9 : 324 0^{\circ} n

36 0^{\circ} n \cdot 9

Multiply the numbers:

2 \cdot 9 = 18

= 324 0^{\circ} n

18 0^{\circ} + 9 θ - 27 θ = 162 0^{\circ} + 324 0^{\circ} n

18 0^{\circ} - 18 θ = 162 0^{\circ} + 324 0^{\circ} n

18 0^{\circ} - 18 θ = 162 0^{\circ} + 324 0^{\circ} n

18 0^{\circ} - 18 θ = 162 0^{\circ} + 324 0^{\circ} n

Move

18 0^{\circ}

to the right side

18 0^{\circ} - 18 θ = 162 0^{\circ} + 324 0^{\circ} n

Subtract

18 0^{\circ}

from both sides

18 0^{\circ} - 18 θ - 18 0^{\circ} = 162 0^{\circ} + 324 0^{\circ} n - 18 0^{\circ}

Simplify

- 18 θ = 144 0^{\circ} + 324 0^{\circ} n

- 18 θ = 144 0^{\circ} + 324 0^{\circ} n

Divide both sides by

- 18

- 18 θ = 144 0^{\circ} + 324 0^{\circ} n

Divide both sides by

- 18

\frac{- 18 θ}{- 18} = \frac{144 0 ^{\circ}}{- 18} + \frac{324 0 ^{\circ} n}{- 18}

Simplify

\frac{- 18 θ}{- 18} = \frac{144 0 ^{\circ}}{- 18} + \frac{324 0 ^{\circ} n}{- 18}

Simplify

\frac{- 18 θ}{- 18} : θ

\frac{- 18 θ}{- 18}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{18 θ}{18}

Divide the numbers:

\frac{18}{18} = 1

= θ

Simplify

\frac{144 0 ^{\circ}}{- 18} + \frac{324 0 ^{\circ} n}{- 18} : - 8 0^{\circ} - 18 0^{\circ} n

\frac{144 0 ^{\circ}}{- 18} + \frac{324 0 ^{\circ} n}{- 18}

\frac{144 0 ^{\circ}}{- 18} = - 8 0^{\circ}

\frac{144 0 ^{\circ}}{- 18}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - 8 0^{\circ}

Cancel the common factor:

2

= - 8 0^{\circ}

= - 8 0^{\circ} + \frac{324 0 ^{\circ} n}{- 18}

\frac{324 0 ^{\circ} n}{- 18} = - 18 0^{\circ} n

\frac{324 0 ^{\circ} n}{- 18}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{324 0 ^{\circ} n}{18}

Divide the numbers:

\frac{18}{18} = 1

= - 18 0^{\circ} n

= - 8 0^{\circ} - 18 0^{\circ} n

θ = - 8 0^{\circ} - 18 0^{\circ} n

θ = - 8 0^{\circ} - 18 0^{\circ} n

θ = - 8 0^{\circ} - 18 0^{\circ} n

θ = - 18 0^{\circ} n + 1 0^{\circ}, θ = - 8 0^{\circ} - 18 0^{\circ} n

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for tan(θ+20)tan(90-3θ)=1 ?
The general solution for tan(θ+20)tan(90-3θ)=1 is θ=-180n+10,θ=-80-180n