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

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

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

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

x = \frac{π}{4} + πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

tan (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 = tan (x)

tan (x) = 1

tan (x) = 1

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn

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

sin (x) = - \frac{12}{13}

2 sin^{2} (x) + 7 sin (x) - 4 = 0

cos (\frac{5 π}{4} + x) + sin (\frac{5 π}{4} - x) = 0

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

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