What is the general solution for sin(θ)csc(3θ-40)=1,calculetan(3θ) ?

The general solution for sin(θ)csc(3θ-40)=1,calculetan(3θ) is No Solution for θ\in\mathbb{R}

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Popular Trigonometry >

sin(θ)csc(3θ-40)=1,calculetan(3θ)

Solution

sin (θ) csc (3 θ - 40) = 1, c a l c u l e tan (3 θ)

Solution

No Solution for θ \in R

Solution steps

sin (θ) csc (3 θ - 40) = 1, c a l c u l e tan (3 θ)

Subtract

1

from both sides

sin (θ) csc (3 θ - 40) - 1 = 0

Express with sin, cos

- 1 + csc (- 40 + 3 θ) sin (θ)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

= - 1 + \frac{1}{sin ( - 40 + 3 θ )} sin (θ)

Simplify

- 1 + \frac{1}{sin ( - 40 + 3 θ )} sin (θ) : \frac{- sin ( - 40 + 3 θ ) + sin ( θ )}{sin ( - 40 + 3 θ )}

- 1 + \frac{1}{sin ( - 40 + 3 θ )} sin (θ)

\frac{1}{sin ( - 40 + 3 θ )} sin (θ) = \frac{sin ( θ )}{sin ( - 40 + 3 θ )}

\frac{1}{sin ( - 40 + 3 θ )} sin (θ)

Multiply fractions:

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

= \frac{1 \cdot sin ( θ )}{sin ( - 40 + 3 θ )}

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ )}{sin ( - 40 + 3 θ )}

= - 1 + \frac{sin ( θ )}{sin ( 3 θ - 40 )}

Convert element to fraction:

1 = \frac{1 sin ( - 40 + 3 θ )}{sin ( - 40 + 3 θ )}

= - \frac{1 \cdot sin ( - 40 + 3 θ )}{sin ( - 40 + 3 θ )} + \frac{sin ( θ )}{sin ( - 40 + 3 θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot sin ( - 40 + 3 θ ) + sin ( θ )}{sin ( - 40 + 3 θ )}

Multiply:

1 \cdot sin (- 40 + 3 θ) = sin (- 40 + 3 θ)

= \frac{- sin ( 3 θ - 40 ) + sin ( θ )}{sin ( - 40 + 3 θ )}

= \frac{- sin ( - 40 + 3 θ ) + sin ( θ )}{sin ( - 40 + 3 θ )}

\frac{- sin ( - 40 + 3 θ ) + sin ( θ )}{sin ( - 40 + 3 θ )} = 0

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

- sin (- 40 + 3 θ) + sin (θ) = 0

Rewrite using trig identities

- sin (- 40 + 3 θ) + sin (θ)

Use the Sum to Product identity:

sin (s) - sin (t) = 2 sin (\frac{s - t}{2}) cos (\frac{s + t}{2})

= 2 sin (\frac{θ - ( - 40 + 3 θ )}{2}) cos (\frac{θ - 40 + 3 θ}{2})

Simplify

2 sin (\frac{θ - ( - 40 + 3 θ )}{2}) cos (\frac{θ - 40 + 3 θ}{2}) : 2 sin (- θ + 20) cos (2 (θ - 10))

2 sin (\frac{θ - ( - 40 + 3 θ )}{2}) cos (\frac{θ - 40 + 3 θ}{2})

\frac{θ - ( - 40 + 3 θ )}{2} = - θ + 20

\frac{θ - ( - 40 + 3 θ )}{2}

Expand

θ - (- 40 + 3 θ) : - 2 θ + 40

θ - (- 40 + 3 θ)

- (- 40 + 3 θ) : 40 - 3 θ

- (- 40 + 3 θ)

Distribute parentheses

= - (- 40) - (3 θ)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= 40 - 3 θ

= θ + 40 - 3 θ

Simplify

θ + 40 - 3 θ : - 2 θ + 40

θ + 40 - 3 θ

Group like terms

= θ - 3 θ + 40

Add similar elements:

θ - 3 θ = - 2 θ

= - 2 θ + 40

= - 2 θ + 40

= \frac{- 2 θ + 40}{2}

Factor

- 2 θ + 40 : 2 (- θ + 20)

- 2 θ + 40

Rewrite as

= - 2 θ + 2 \cdot 20

Factor out common term

2

= 2 (- θ + 20)

= \frac{2 ( - θ + 20 )}{2}

Divide the numbers:

\frac{2}{2} = 1

= - θ + 20

= 2 sin ((- θ + 20)) cos (\frac{θ + 3 θ - 40}{2})

\frac{θ - 40 + 3 θ}{2} = 2 (θ - 10)

\frac{θ - 40 + 3 θ}{2}

θ - 40 + 3 θ = 4 θ - 40

θ - 40 + 3 θ

Group like terms

= θ + 3 θ - 40

Add similar elements:

θ + 3 θ = 4 θ

= 4 θ - 40

= \frac{4 θ - 40}{2}

Factor

4 θ - 40 : 4 (θ - 10)

4 θ - 40

Rewrite as

= 4 θ - 4 \cdot 10

Factor out common term

4

= 4 (θ - 10)

= \frac{4 ( θ - 10 )}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2 (θ - 10)

= 2 sin ((- θ + 20)) cos (2 (θ - 10))

Remove parentheses:

(- a) = - a

= 2 sin (- θ + 20) cos (2 (θ - 10))

= 2 sin (- θ + 20) cos (2 (θ - 10))

2 cos ((- 10 + θ) \cdot 2) sin (20 - θ) = 0

Solving each part separately

cos ((- 10 + θ) \cdot 2) = 0 or sin (20 - θ) = 0

cos ((- 10 + θ) \cdot 2) = 0, c a l c u l e tan (3 θ) :

No Solution

cos ((- 10 + θ) \cdot 2) = 0, c a l c u l e tan (3 θ)

General solutions for

cos ((- 10 + θ) 2) = 0

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}

(- 10 + θ) \cdot 2 = \frac{π}{2} + 2 πn, (- 10 + θ) \cdot 2 = \frac{3 π}{2} + 2 πn

(- 10 + θ) \cdot 2 = \frac{π}{2} + 2 πn, (- 10 + θ) \cdot 2 = \frac{3 π}{2} + 2 πn

Solve

(- 10 + θ) 2 = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn + 10

(- 10 + θ) \cdot 2 = \frac{π}{2} + 2 πn

Divide both sides by

2

(- 10 + θ) \cdot 2 = \frac{π}{2} + 2 πn

Divide both sides by

2

\frac{( - 10 + θ ) \cdot 2}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{( - 10 + θ ) \cdot 2}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{( - 10 + θ ) \cdot 2}{2} : - 10 + θ

\frac{( - 10 + θ ) \cdot 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 10 + θ

Simplify

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

- 10 + θ = \frac{π}{4} + πn

- 10 + θ = \frac{π}{4} + πn

- 10 + θ = \frac{π}{4} + πn

Move

10

to the right side

- 10 + θ = \frac{π}{4} + πn

Add

10

to both sides

- 10 + θ + 10 = \frac{π}{4} + πn + 10

Simplify

θ = \frac{π}{4} + πn + 10

θ = \frac{π}{4} + πn + 10

Solve

(- 10 + θ) 2 = \frac{3 π}{2} + 2 πn : θ = \frac{3 π}{4} + πn + 10

(- 10 + θ) \cdot 2 = \frac{3 π}{2} + 2 πn

Divide both sides by

2

(- 10 + θ) \cdot 2 = \frac{3 π}{2} + 2 πn

Divide both sides by

2

\frac{( - 10 + θ ) \cdot 2}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{( - 10 + θ ) \cdot 2}{2} = \frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

Simplify

\frac{( - 10 + θ ) \cdot 2}{2} : - 10 + θ

\frac{( - 10 + θ ) \cdot 2}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 10 + θ

Simplify

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

- 10 + θ = \frac{3 π}{4} + πn

- 10 + θ = \frac{3 π}{4} + πn

- 10 + θ = \frac{3 π}{4} + πn

Move

10

to the right side

- 10 + θ = \frac{3 π}{4} + πn

Add

10

to both sides

- 10 + θ + 10 = \frac{3 π}{4} + πn + 10

Simplify

θ = \frac{3 π}{4} + πn + 10

θ = \frac{3 π}{4} + πn + 10

θ = \frac{π}{4} + πn + 10, θ = \frac{3 π}{4} + πn + 10

Solutions for the range

c a l c u l e tan (3 θ)

No Solution

sin (20 - θ) = 0, c a l c u l e tan (3 θ) :

No Solution

sin (20 - θ) = 0, c a l c u l e tan (3 θ)

General solutions for

sin (20 - 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}

20 - θ = 0 + 2 πn, 20 - θ = π + 2 πn

20 - θ = 0 + 2 πn, 20 - θ = π + 2 πn

Solve

20 - θ = 0 + 2 πn : θ = - 2 πn + 20

20 - θ = 0 + 2 πn

0 + 2 πn = 2 πn

20 - θ = 2 πn

Move

20

to the right side

20 - θ = 2 πn

Subtract

20

from both sides

20 - θ - 20 = 2 πn - 20

Simplify

- θ = 2 πn - 20

- θ = 2 πn - 20

Divide both sides by

- 1

- θ = 2 πn - 20

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

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

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{2 πn}{- 1} - \frac{20}{- 1} : - 2 πn + 20

\frac{2 πn}{- 1} - \frac{20}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - 2 πn - \frac{20}{- 1}

\frac{20}{- 1} = - 20

\frac{20}{- 1}

Apply the fraction rule:

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

= - \frac{20}{1}

Apply rule

\frac{a}{1} = a

= - 20

= - 2 πn - (- 20)

Apply rule

- (- a) = a

= - 2 πn + 20

θ = - 2 πn + 20

θ = - 2 πn + 20

θ = - 2 πn + 20

Solve

20 - θ = π + 2 πn : θ = - π + 20 - 2 πn

20 - θ = π + 2 πn

Move

20

to the right side

20 - θ = π + 2 πn

Subtract

20

from both sides

20 - θ - 20 = π + 2 πn - 20

Simplify

- θ = π + 2 πn - 20

- θ = π + 2 πn - 20

Divide both sides by

- 1

- θ = π + 2 πn - 20

Divide both sides by

- 1

\frac{- θ}{- 1} = \frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- θ}{- 1} = \frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1}

Simplify

\frac{- θ}{- 1} : θ

\frac{- θ}{- 1}

Apply the fraction rule:

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

= \frac{θ}{1}

Apply rule

\frac{a}{1} = a

= θ

Simplify

\frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1} : - π + 20 - 2 πn

\frac{π}{- 1} + \frac{2 πn}{- 1} - \frac{20}{- 1}

Group like terms

= \frac{π}{- 1} - \frac{20}{- 1} + \frac{2 πn}{- 1}

\frac{π}{- 1} = - π

\frac{π}{- 1}

Apply the fraction rule:

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

= - \frac{π}{1}

Apply rule

\frac{a}{1} = a

= - π

= - π - \frac{20}{- 1} + \frac{2 πn}{- 1}

\frac{20}{- 1} = - 20

\frac{20}{- 1}

Apply the fraction rule:

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

= - \frac{20}{1}

Apply rule

\frac{a}{1} = a

= - 20

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

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

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - π - (- 20) - 2 πn

Apply rule

- (- a) = a

= - π + 20 - 2 πn

θ = - π + 20 - 2 πn

θ = - π + 20 - 2 πn

θ = - π + 20 - 2 πn

θ = - 2 πn + 20, θ = - π + 20 - 2 πn

Solutions for the range

c a l c u l e tan (3 θ)

No Solution

Combine all the solutions

No Solution for θ \in R

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)csc(3θ-40)=1,calculetan(3θ) ?
The general solution for sin(θ)csc(3θ-40)=1,calculetan(3θ) is No Solution for θ\in\mathbb{R}