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

The general solution for csc^2(θ)-2cot(θ)=0 is θ= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

csc^2(θ)-2cot(θ)=0

Solution

csc^{2} (θ) - 2 cot (θ) = 0

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

csc^{2} (θ) - 2 cot (θ) = 0

Rewrite using trig identities

csc^{2} (θ) - 2 cot (θ)

Use the Pythagorean identity:

csc^{2} (x) = 1 + cot^{2} (x)

= 1 + cot^{2} (θ) - 2 cot (θ)

1 + cot^{2} (θ) - 2 cot (θ) = 0

Solve by substitution

1 + cot^{2} (θ) - 2 cot (θ) = 0

Let:

cot (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

u = 1

The solution to the quadratic equation is:

u = 1

Substitute back

u = cot (θ)

cot (θ) = 1

cot (θ) = 1

cot (θ) = 1 : θ = \frac{π}{4} + πn

cot (θ) = 1

General solutions for

cot (θ) = 1

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

Popular Examples

cos(θ)=0.3,0<= θ<= 2pi

cos (θ) = 0.3, 0 \leq θ \leq 2 π

cos(3x)=-(sqrt(2))/2

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

2cos(x)+1=2

2 cos (x) + 1 = 2

sin(θ)= 5/9

sin (θ) = \frac{5}{9}

1+sin(x)=2sin^2(x)

1 + sin (x) = 2 sin^{2} (x)

Frequently Asked Questions (FAQ)

What is the general solution for csc^2(θ)-2cot(θ)=0 ?
The general solution for csc^2(θ)-2cot(θ)=0 is θ= pi/4+pin