What is the general solution for 3+sin(θ)=2csc(θ) ?

The general solution for 3+sin(θ)=2csc(θ) is θ=0.59626…+2pin,θ=pi-0.59626…+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

3+sin(θ)=2csc(θ)

Solution

3 + sin (θ) = 2 csc (θ)

Solution

θ = 0.59626 \dots + 2 πn, θ = π - 0.59626 \dots + 2 πn

Degrees

θ = 34.16325 \dots^{\circ} + 36 0^{\circ} n, θ = 145.83674 \dots^{\circ} + 36 0^{\circ} n

Solution steps

3 + sin (θ) = 2 csc (θ)

Subtract

2 csc (θ)

from both sides

3 + sin (θ) - 2 csc (θ) = 0

Rewrite using trig identities

3 + sin (θ) - 2 csc (θ)

Use the basic trigonometric identity:

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

= 3 + \frac{1}{csc ( θ )} - 2 csc (θ)

3 + \frac{1}{csc ( θ )} - 2 csc (θ) = 0

Solve by substitution

3 + \frac{1}{csc ( θ )} - 2 csc (θ) = 0

Let:

csc (θ) = u

3 + \frac{1}{u} - 2 u = 0

3 + \frac{1}{u} - 2 u = 0 : u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

3 + \frac{1}{u} - 2 u = 0

Multiply both sides by

u

3 + \frac{1}{u} - 2 u = 0

Multiply both sides by

u

3 u + \frac{1}{u} u - 2 uu = 0 \cdot u

Simplify

3 u + \frac{1}{u} u - 2 uu = 0 \cdot u

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

- 2 uu : - 2 u^{2}

- 2 uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 2 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

3 u + 1 - 2 u^{2} = 0

3 u + 1 - 2 u^{2} = 0

3 u + 1 - 2 u^{2} = 0

Solve

3 u + 1 - 2 u^{2} = 0 : u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

3 u + 1 - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + 3 u + 1 = 0

Solve with the quadratic formula

- 2 u^{2} + 3 u + 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

3^{2} - 4 (- 2) \cdot 1 = 17

3^{2} - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

Add the numbers:

9 + 8 = 17

= 17

u_{1, 2} = \frac{- 3 \pm 17}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 3 + 17}{2 ( - 2 )}, u_{2} = \frac{- 3 - 17}{2 ( - 2 )}

u = \frac{- 3 + 17}{2 ( - 2 )} : - \frac{- 3 + 17}{4}

\frac{- 3 + 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 + 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 + 17}{- 4}

Apply the fraction rule:

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

= - \frac{- 3 + 17}{4}

u = \frac{- 3 - 17}{2 ( - 2 )} : \frac{3 + 17}{4}

\frac{- 3 - 17}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 3 - 17}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 - 17}{- 4}

Apply the fraction rule:

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

- 3 - 17 = - (3 + 17)

= \frac{3 + 17}{4}

The solutions to the quadratic equation are:

u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

3 + \frac{1}{u} - 2 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - \frac{- 3 + 17}{4}, u = \frac{3 + 17}{4}

Substitute back

u = csc (θ)

csc (θ) = - \frac{- 3 + 17}{4}, csc (θ) = \frac{3 + 17}{4}

csc (θ) = - \frac{- 3 + 17}{4}, csc (θ) = \frac{3 + 17}{4}

csc (θ) = - \frac{- 3 + 17}{4} :

No Solution

csc (θ) = - \frac{- 3 + 17}{4}

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

csc (θ) = \frac{3 + 17}{4} : θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

csc (θ) = \frac{3 + 17}{4}

Apply trig inverse properties

csc (θ) = \frac{3 + 17}{4}

General solutions for

csc (θ) = \frac{3 + 17}{4}

csc (x) = a \Rightarrow x = arccsc (a) + 2 πn, x = π - arccsc (a) + 2 πn

θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

Combine all the solutions

θ = arccsc (\frac{3 + 17}{4}) + 2 πn, θ = π - arccsc (\frac{3 + 17}{4}) + 2 πn

Show solutions in decimal form

θ = 0.59626 \dots + 2 πn, θ = π - 0.59626 \dots + 2 πn

Graph

Popular Examples

10sin^2(x)-3sin(x)-1=0

cos(2x)-cos(pi/2+x)=1

cos(3x)-cos(7x)=0

2sin(x-1)=0

tan(x)=0.42

Frequently Asked Questions (FAQ)

What is the general solution for 3+sin(θ)=2csc(θ) ?
The general solution for 3+sin(θ)=2csc(θ) is θ=0.59626…+2pin,θ=pi-0.59626…+2pin