What is the general solution for 5cot(θ)-2=3cot(θ)-2 ?

The general solution for 5cot(θ)-2=3cot(θ)-2 is θ= pi/2+pin

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5cot(θ)-2=3cot(θ)-2

5 cot (θ) - 2 = 3 cot (θ) - 2

θ = \frac{π}{2} + πn

Degrees

θ = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

5 cot (θ) - 2 = 3 cot (θ) - 2

Solve by substitution

5 cot (θ) - 2 = 3 cot (θ) - 2

Let:

cot (θ) = u

5 u - 2 = 3 u - 2

5 u - 2 = 3 u - 2 : u = 0

5 u - 2 = 3 u - 2

Add

2

to both sides

5 u - 2 + 2 = 3 u - 2 + 2

Simplify

5 u = 3 u

Move

3 u

to the left side

5 u = 3 u

Subtract

3 u

from both sides

5 u - 3 u = 3 u - 3 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

\frac{sin ( x )}{sin ( 2 x )} = 0

5 cos^{2} (x) = \frac{1}{2 ( 1 + cos ^{2} ( x ) )}

2 sin^{2} (\frac{x}{2}) - 1 = sin (\frac{x}{2})

4.7 2^{2} = 3. 3^{2} + 3. 3^{2} - 2 (3.3) (3.3) cos (x)

2 sin (x - 3 0^{\circ}) = cos (x - 6 0^{\circ})

What is the general solution for 5cot(θ)-2=3cot(θ)-2 ?
The general solution for 5cot(θ)-2=3cot(θ)-2 is θ= pi/2+pin