What is the general solution for 4(cot(θ)+1)=2(cot(θ)+2) ?

The general solution for 4(cot(θ)+1)=2(cot(θ)+2) is θ= pi/2+pin

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Popular >

4(cot(θ)+1)=2(cot(θ)+2)

Solution

4 (cot (θ) + 1) = 2 (cot (θ) + 2)

Solution

θ = \frac{π}{2} + πn

Degrees

θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

4 (cot (θ) + 1) = 2 (cot (θ) + 2)

Solve by substitution

4 (cot (θ) + 1) = 2 (cot (θ) + 2)

Let:

cot (θ) = u

4 (u + 1) = 2 (u + 2)

4 (u + 1) = 2 (u + 2) : u = 0

4 (u + 1) = 2 (u + 2)

Expand

4 (u + 1) : 4 u + 4

4 (u + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = 4, b = u, c = 1

= 4 u + 4 \cdot 1

Multiply the numbers:

4 \cdot 1 = 4

= 4 u + 4

Expand

2 (u + 2) : 2 u + 4

2 (u + 2)

Apply the distributive law:

a (b + c) = ab + a c

a = 2, b = u, c = 2

= 2 u + 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 2 u + 4

4 u + 4 = 2 u + 4

Move

4

to the right side

4 u + 4 = 2 u + 4

Subtract

4

from both sides

4 u + 4 - 4 = 2 u + 4 - 4

Simplify

4 u = 2 u

4 u = 2 u

Move

2 u

to the left side

4 u = 2 u

Subtract

2 u

from both sides

4 u - 2 u = 2 u - 2 u

Simplify

2 u = 0

2 u = 0

Divide both sides by

2

2 u = 0

Divide both sides by

2

\frac{2 u}{2} = \frac{0}{2}

Simplify

u = 0

u = 0

Substitute back

u = cot (θ)

cot (θ) = 0

cot (θ) = 0

cot (θ) = 0 : θ = \frac{π}{2} + πn

cot (θ) = 0

General solutions for

cot (θ) = 0

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}{2} + πn

θ = \frac{π}{2} + πn

Combine all the solutions

θ = \frac{π}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4(cot(θ)+1)=2(cot(θ)+2) ?
The general solution for 4(cot(θ)+1)=2(cot(θ)+2) is θ= pi/2+pin