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

The general solution for cot^2(x)-2cot(x)+1=0 is x= pi/4+pin

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

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

x = \frac{π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cot (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 (x)

cot (x) = 1

cot (x) = 1

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

General solutions for

cot (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

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn

3 sin (x) - 2 = 0

4 cos^{2} (x) - 2 sin (x) + 1 = 0

tan (x) = \frac{3 3}{5}

tan (x) + 1 = - \frac{1}{3} - \frac{1}{3} cot (x)

6 cos^{2} (x) + 7 cos (x) + 2 = 0

What is the general solution for cot^2(x)-2cot(x)+1=0 ?
The general solution for cot^2(x)-2cot(x)+1=0 is x= pi/4+pin