What is the general solution for cot^2(3x)-2cot(3x)+1=0 ?

The general solution for cot^2(3x)-2cot(3x)+1=0 is x= pi/(12)+(pin)/3

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cot^2(3x)-2cot(3x)+1=0

Solution

cot^{2} (3 x) - 2 cot (3 x) + 1 = 0

Solution

x = \frac{π}{12} + \frac{πn}{3}

Degrees

x = 1 5^{\circ} + 6 0^{\circ} n

Solution steps

cot^{2} (3 x) - 2 cot (3 x) + 1 = 0

Solve by substitution

cot^{2} (3 x) - 2 cot (3 x) + 1 = 0

Let:

cot (3 x) = u

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

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

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 (3 x)

cot (3 x) = 1

cot (3 x) = 1

cot (3 x) = 1 : x = \frac{π}{12} + \frac{πn}{3}

cot (3 x) = 1

General solutions for

cot (3 x) = 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

3 x = \frac{π}{4} + πn

3 x = \frac{π}{4} + πn

Solve

3 x = \frac{π}{4} + πn : x = \frac{π}{12} + \frac{πn}{3}

3 x = \frac{π}{4} + πn

Divide both sides by

3

3 x = \frac{π}{4} + πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{4}}{3} + \frac{πn}{3} : \frac{π}{12} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

Combine all the solutions

x = \frac{π}{12} + \frac{πn}{3}

Popular Examples

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- \frac{tanh ( n x )}{cosh ( n x )} = 0

sin(φ)=0

sin (φ) = 0

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sin^{2} (x) = 1 - tan^{2} (x), 0 \leq x \leq \frac{π}{2}

sin(0.3x)=1

sin (0.3 x) = 1

22.89=cos(a)

22.89 = cos (a)

Frequently Asked Questions (FAQ)

What is the general solution for cot^2(3x)-2cot(3x)+1=0 ?
The general solution for cot^2(3x)-2cot(3x)+1=0 is x= pi/(12)+(pin)/3