What is the general solution for csc^2(θ)-2csc(θ)=0 ?

The general solution for csc^2(θ)-2csc(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

csc^2(θ)-2csc(θ)=0

Solution

csc^{2} (θ) - 2 csc (θ) = 0

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

csc^{2} (θ) - 2 csc (θ) = 0

Solve by substitution

csc^{2} (θ) - 2 csc (θ) = 0

Let:

csc (θ) = u

u^{2} - 2 u = 0

u^{2} - 2 u = 0 : u = 2, u = 0

u^{2} - 2 u = 0

Solve with the quadratic formula

u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 1 \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

u = \frac{- ( - 2 ) + 2}{2 \cdot 1} : 2

\frac{- ( - 2 ) + 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 + 2}{2 \cdot 1}

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 2 ) - 2}{2 \cdot 1} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{2 - 2}{2 \cdot 1}

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = 2, u = 0

Substitute back

u = csc (θ)

csc (θ) = 2, csc (θ) = 0

csc (θ) = 2, csc (θ) = 0

csc (θ) = 2 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

csc (θ) = 2

General solutions for

csc (θ) = 2

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

csc (θ) = 0 :

No Solution

csc (θ) = 0

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

No Solution

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Popular Examples

cos(θ)=-7/25

cos (θ) = - \frac{7}{25}

3cos(x)+3sin(x)tan(x)=6

3 cos (x) + 3 sin (x) tan (x) = 6

tan^2(2x)-1=0

tan^{2} (2 x) - 1 = 0

3sin(4x)=-6sin(2x)

3 sin (4 x) = - 6 sin (2 x)

2sin(2θ)-sqrt(3)=0

2 sin (2 θ) - 3 = 0

Frequently Asked Questions (FAQ)

What is the general solution for csc^2(θ)-2csc(θ)=0 ?
The general solution for csc^2(θ)-2csc(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin