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

The general solution for sin(x^2-2x+1)=1 is x=(4+sqrt(32pin+8pi))/4 ,x=(4-sqrt(32pin+8pi))/4

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sin(x^2-2x+1)=1

Solution

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

Solution

x = \frac{4 + 32 πn + 8 π}{4}, x = \frac{4 - 32 πn + 8 π}{4}

Degrees

x = 57.29577 \dots^{\circ} + 160.57117 \dots^{\circ} n, x = 57.29577 \dots^{\circ} - 160.57117 \dots^{\circ} n

Solution steps

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

General solutions for

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

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x^{2} - 2 x + 1 = \frac{π}{2} + 2 πn

x^{2} - 2 x + 1 = \frac{π}{2} + 2 πn

Solve

x^{2} - 2 x + 1 = \frac{π}{2} + 2 πn : x = \frac{4 + 32 πn + 8 π}{4}, x = \frac{4 - 32 πn + 8 π}{4}

x^{2} - 2 x + 1 = \frac{π}{2} + 2 πn

Multiply both sides by

2

x^{2} - 2 x + 1 = \frac{π}{2} + 2 πn

Multiply both sides by

2

x^{2} \cdot 2 - 2 x \cdot 2 + 1 \cdot 2 = \frac{π}{2} \cdot 2 + 2 πn \cdot 2

Simplify

2 x^{2} - 4 x + 2 = π + 4 πn

2 x^{2} - 4 x + 2 = π + 4 πn

Move

4 πn

to the left side

2 x^{2} - 4 x + 2 = π + 4 πn

Subtract

4 πn

from both sides

2 x^{2} - 4 x + 2 - 4 πn = π + 4 πn - 4 πn

Simplify

2 x^{2} - 4 x + 2 - 4 πn = π

2 x^{2} - 4 x + 2 - 4 πn = π

Move

π

to the left side

2 x^{2} - 4 x + 2 - 4 πn = π

Subtract

π

from both sides

2 x^{2} - 4 x + 2 - 4 πn - π = π - π

Simplify

2 x^{2} - 4 x + 2 - 4 πn - π = 0

2 x^{2} - 4 x + 2 - 4 πn - π = 0

Solve with the quadratic formula

2 x^{2} - 4 x + 2 - 4 πn - π = 0

Quadratic Equation Formula:

For

a = 2, b = - 4, c = 2 - 4 πn - π

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( 2 - 4 πn - π )}{2 \cdot 2}

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( 2 - 4 πn - π )}{2 \cdot 2}

Simplify

(- 4)^{2} - 4 \cdot 2 (2 - 4 πn - π) : 32 πn + 8 π

(- 4)^{2} - 4 \cdot 2 (2 - 4 πn - π)

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 2 (- 4 πn - π + 2)

Refine

= 16 - 8 (- 4 πn - π + 2)

Expand

16 - 8 (2 - 4 πn - π) : 32 πn + 8 π

16 - 8 (2 - 4 πn - π)

Expand

- 8 (2 - 4 πn - π) : - 16 + 32 πn + 8 π

- 8 (2 - 4 πn - π)

Distribute parentheses

= (- 8) \cdot 2 + (- 8) (- 4 πn) + (- 8) (- π)

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

= - 8 \cdot 2 + 8 \cdot 4 πn + 8 π

Simplify

- 8 \cdot 2 + 8 \cdot 4 πn + 8 π : - 16 + 32 πn + 8 π

- 8 \cdot 2 + 8 \cdot 4 πn + 8 π

Multiply the numbers:

8 \cdot 2 = 16

= - 16 + 8 \cdot 4 πn + 8 π

Multiply the numbers:

8 \cdot 4 = 32

= - 16 + 32 πn + 8 π

= - 16 + 32 πn + 8 π

= 16 - 16 + 32 πn + 8 π

16 - 16 = 0

= 32 πn + 8 π

= 32 πn + 8 π

x_{1, 2} = \frac{- ( - 4 ) \pm 32 πn + 8 π}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 32 πn + 8 π}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 32 πn + 8 π}{2 \cdot 2}

x = \frac{- ( - 4 ) + 32 πn + 8 π}{2 \cdot 2} : \frac{4 + 32 πn + 8 π}{4}

\frac{- ( - 4 ) + 32 πn + 8 π}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4 + 32 πn + 8 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 + 32 πn + 8 π}{4}

x = \frac{- ( - 4 ) - 32 πn + 8 π}{2 \cdot 2} : \frac{4 - 32 πn + 8 π}{4}

\frac{- ( - 4 ) - 32 πn + 8 π}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{4 - 32 πn + 8 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 - 32 πn + 8 π}{4}

The solutions to the quadratic equation are:

x = \frac{4 + 32 πn + 8 π}{4}, x = \frac{4 - 32 πn + 8 π}{4}

x = \frac{4 + 32 πn + 8 π}{4}, x = \frac{4 - 32 πn + 8 π}{4}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x^2-2x+1)=1 ?
The general solution for sin(x^2-2x+1)=1 is x=(4+sqrt(32pin+8pi))/4 ,x=(4-sqrt(32pin+8pi))/4