zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

x = 1 + 2 πn, x = 1 - 2 πn, x = 1 + π (2 n + 1), x = 1 - π (2 n + 1)

+1

度数

x = 57.29577 \dots^{\circ} + 143.61922 \dots^{\circ} n, x = 57.29577 \dots^{\circ} - 143.61922 \dots^{\circ} n, x = 57.29577 \dots^{\circ} + 175.89690 \dots^{\circ} n, x = 57.29577 \dots^{\circ} - 175.89690 \dots^{\circ} n

求解步骤

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

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

的通解

sin (x)

周期表（周期为

2 πn

"）：

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 = 0 + 2 πn, x^{2} - 2 x + 1 = π + 2 πn

x^{2} - 2 x + 1 = 0 + 2 πn, x^{2} - 2 x + 1 = π + 2 πn

解

x^{2} - 2 x + 1 = 0 + 2 πn : x = 1 + 2 πn, x = 1 - 2 πn

x^{2} - 2 x + 1 = 0 + 2 πn

展开

0 + 2 πn : 2 πn

0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

x^{2} - 2 x + 1 = 2 πn

将

2 πn

para o lado esquerdo

x^{2} - 2 x + 1 = 2 πn

两边减去

2 πn

x^{2} - 2 x + 1 - 2 πn = 2 πn - 2 πn

化简

x^{2} - 2 x + 1 - 2 πn = 0

x^{2} - 2 x + 1 - 2 πn = 0

使用求根公式求解

x^{2} - 2 x + 1 - 2 πn = 0

二次方程求根公式:

若

a = 1, b = - 2, c = 1 - 2 πn

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

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

化简

(- 2)^{2} - 4 \cdot 1 \cdot (1 - 2 πn) : 22 πn

(- 2)^{2} - 4 \cdot 1 \cdot (1 - 2 πn)

使用指数法则:

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

若

n

是偶数

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

= 2^{2} - 4 \cdot 1 \cdot (- 2 πn + 1)

整理后得

= 4 - 4 (- 2 πn + 1)

乘开

4 - 4 (1 - 2 πn) : 8 πn

4 - 4 (1 - 2 πn)

乘开

- 4 (1 - 2 πn) : - 4 + 8 πn

- 4 (1 - 2 πn)

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 1, c = 2 πn

= - 4 \cdot 1 - (- 4) \cdot 2 πn

使用加减运算法则

- (- a) = a

= - 4 \cdot 1 + 4 \cdot 2 πn

化简

- 4 \cdot 1 + 4 \cdot 2 πn : - 4 + 8 πn

- 4 \cdot 1 + 4 \cdot 2 πn

数字相乘:

4 \cdot 1 = 4

= - 4 + 4 \cdot 2 πn

数字相乘:

4 \cdot 2 = 8

= - 4 + 8 πn

= - 4 + 8 πn

= 4 - 4 + 8 πn

4 - 4 = 0

= 8 πn

= 8 πn

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 8 πn

8 = 22

8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

= 22 πn

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

将解分隔开

x_{1} = \frac{- ( - 2 ) + 2 2 πn}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 2 πn}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 2 πn}{2 \cdot 1} : 1 + 2 πn

\frac{- ( - 2 ) + 2 2 πn}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 + 2 2 πn}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 2 πn}{2}

分解

2 + 22 πn : 2 (1 + 2 nπ)

2 + 22 πn

改写为

= 2 \cdot 1 + 22 nπ

因式分解出通项

2

= 2 (1 + 2 nπ)

= \frac{2 ( 1 + 2 nπ )}{2}

数字相除:

\frac{2}{2} = 1

= 1 + 2 πn

x = \frac{- ( - 2 ) - 2 2 πn}{2 \cdot 1} : 1 - 2 πn

\frac{- ( - 2 ) - 2 2 πn}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 - 2 2 πn}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 2 πn}{2}

分解

2 - 22 πn : 2 (1 - 2 nπ)

2 - 22 πn

改写为

= 2 \cdot 1 - 22 nπ

因式分解出通项

2

= 2 (1 - 2 nπ)

= \frac{2 ( 1 - 2 nπ )}{2}

数字相除:

\frac{2}{2} = 1

= 1 - 2 πn

二次方程组的解是:

x = 1 + 2 πn, x = 1 - 2 πn

解

x^{2} - 2 x + 1 = π + 2 πn : x = 1 + π (2 n + 1), x = 1 - π (2 n + 1)

x^{2} - 2 x + 1 = π + 2 πn

将

2 πn

para o lado esquerdo

x^{2} - 2 x + 1 = π + 2 πn

两边减去

2 πn

x^{2} - 2 x + 1 - 2 πn = π + 2 πn - 2 πn

化简

x^{2} - 2 x + 1 - 2 πn = π

x^{2} - 2 x + 1 - 2 πn = π

将

π

para o lado esquerdo

x^{2} - 2 x + 1 - 2 πn = π

两边减去

π

x^{2} - 2 x + 1 - 2 πn - π = π - π

化简

x^{2} - 2 x + 1 - 2 πn - π = 0

x^{2} - 2 x + 1 - 2 πn - π = 0

使用求根公式求解

x^{2} - 2 x + 1 - 2 πn - π = 0

二次方程求根公式:

若

a = 1, b = - 2, c = 1 - 2 πn - π

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

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

化简

(- 2)^{2} - 4 \cdot 1 \cdot (1 - 2 πn - π) : 2 π (1 + 2 n)

(- 2)^{2} - 4 \cdot 1 \cdot (1 - 2 πn - π)

使用指数法则:

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

若

n

是偶数

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

= 2^{2} - 4 \cdot 1 \cdot (- 2 πn - π + 1)

数字相乘:

4 \cdot 1 = 4

= 2^{2} - 4 (- 2 πn - π + 1)

分解

2^{2} - 4 (1 - 2 πn - π) : 4 π (1 + 2 n)

2^{2} - 4 (1 - 2 πn - π)

改写为

= 4 \cdot 1 - 4 (1 - π - 2 nπ)

因式分解出通项

4

= 4 (1 - (1 - π - 2 nπ))

分解

- (- π - 2 πn + 1) + 1 : π (1 + 2 n)

1 - (1 - π - 2 nπ)

= 1 - (1 - π - 2 πn)

- (1 - π - 2 nπ) : - 1 + π + 2 nπ

- (1 - π - 2 nπ)

打开括号

= - (1) - (- π) - (- 2 nπ)

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + π + 2 nπ

= 1 - 1 + π + 2 nπ

1 - 1 = 0

= π + 2 πn

因式分解出通项

π

= π (1 + 2 n)

= 4 π (2 n + 1)

= 4 π (1 + 2 n)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 4 π (2 n + 1)

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 π (2 n + 1)

x_{1, 2} = \frac{- ( - 2 ) \pm 2 π ( 1 + 2 n )}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - 2 ) + 2 π ( 1 + 2 n )}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 π ( 1 + 2 n )}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 π ( 1 + 2 n )}{2 \cdot 1} : 1 + π (2 n + 1)

\frac{- ( - 2 ) + 2 π ( 1 + 2 n )}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 + 2 π ( 1 + 2 n )}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 π ( 2 n + 1 )}{2}

分解

2 + 2 π (1 + 2 n) : 2 (1 + (1 + 2 n) π)

2 + 2 π (1 + 2 n)

改写为

= 2 \cdot 1 + 2 (1 + 2 n) π

因式分解出通项

2

= 2 (1 + (1 + 2 n) π)

= \frac{2 ( 1 + ( 1 + 2 n ) π )}{2}

数字相除:

\frac{2}{2} = 1

= 1 + π (2 n + 1)

x = \frac{- ( - 2 ) - 2 π ( 1 + 2 n )}{2 \cdot 1} : 1 - π (2 n + 1)

\frac{- ( - 2 ) - 2 π ( 1 + 2 n )}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 - 2 π ( 1 + 2 n )}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 π ( 2 n + 1 )}{2}

分解

2 - 2 π (1 + 2 n) : 2 (1 - (1 + 2 n) π)

2 - 2 π (1 + 2 n)

改写为

= 2 \cdot 1 - 2 (1 + 2 n) π

因式分解出通项

2

= 2 (1 - (1 + 2 n) π)

= \frac{2 ( 1 - ( 1 + 2 n ) π )}{2}

数字相除:

\frac{2}{2} = 1

= 1 - π (2 n + 1)

二次方程组的解是:

x = 1 + π (2 n + 1), x = 1 - π (2 n + 1)

x = 1 + 2 πn, x = 1 - 2 πn, x = 1 + π (2 n + 1), x = 1 - π (2 n + 1)

作图

流行的例子

cos (θ) = - 3

\frac{1}{cos ( x )} = 2

2tan^2(x)=-3sec(x)

2 tan^{2} (x) = - 3 sec (x)

tan (θ) + 1 = 2

3 sec (θ) + 1 = 7