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受欢迎的三角函数 >

sin(x^2)=sin(x)

解答

sin (x^{2}) = sin (x)

解答

x = \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{1 + 1 + 16 πn}{2}, x = \frac{1 - 1 + 16 πn}{2}, x = \frac{1 + 1 - 4 ( - 4 πn - 2 π )}{2}, x = \frac{1 - 1 - 4 ( - 4 πn - 2 π )}{2}

+1

度数

x = - 28.64788 \dots^{\circ} + 228.88185 \dots^{\circ} n, x = - 28.64788 \dots^{\circ} - 228.88185 \dots^{\circ} n, x = - 28.64788 \dots^{\circ} + 270.20988 \dots^{\circ} n, x = - 28.64788 \dots^{\circ} - 270.20988 \dots^{\circ} n, x = 28.64788 \dots^{\circ} + 205.11865 \dots^{\circ} n, x = 28.64788 \dots^{\circ} - 205.11865 \dots^{\circ} n, x = 28.64788 \dots^{\circ} + 250.39996 \dots^{\circ} n, x = 28.64788 \dots^{\circ} - 250.39996 \dots^{\circ} n

求解步骤

sin (x^{2}) = sin (x)

两边减去

sin (x)

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

使用三角恒等式改写

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

使用和差化积恒等式:

sin (s) - sin (t) = 2 sin (\frac{s - t}{2}) cos (\frac{s + t}{2})

= 2 sin (\frac{x ^{2} - x}{2}) cos (\frac{x ^{2} + x}{2})

2 cos (\frac{x + x ^{2}}{2}) sin (\frac{- x + x ^{2}}{2}) = 0

分别求解每个部分

cos (\frac{x + x ^{2}}{2}) = 0 or sin (\frac{- x + x ^{2}}{2}) = 0

cos (\frac{x + x ^{2}}{2}) = 0 : x = \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}

cos (\frac{x + x ^{2}}{2}) = 0

cos (\frac{x + x ^{2}}{2}) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

\frac{x + x ^{2}}{2} = \frac{π}{2} + 2 πn, \frac{x + x ^{2}}{2} = \frac{3 π}{2} + 2 πn

\frac{x + x ^{2}}{2} = \frac{π}{2} + 2 πn, \frac{x + x ^{2}}{2} = \frac{3 π}{2} + 2 πn

解

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

\frac{x + x ^{2}}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

\frac{x + x ^{2}}{2} = \frac{π}{2} + 2 πn

在两边乘以

2

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

化简

x + x^{2} = π + 4 πn

x + x^{2} = π + 4 πn

将

4 πn

para o lado esquerdo

x + x^{2} = π + 4 πn

两边减去

4 πn

x + x^{2} - 4 πn = π + 4 πn - 4 πn

化简

x + x^{2} - 4 πn = π

x + x^{2} - 4 πn = π

将

π

para o lado esquerdo

x + x^{2} - 4 πn = π

两边减去

π

x + x^{2} - 4 πn - π = π - π

化简

x + x^{2} - 4 πn - π = 0

x + x^{2} - 4 πn - π = 0

改写成标准形式

a x^{2} + b x + c = 0

x^{2} + x - 4 πn - π = 0

使用求根公式求解

x^{2} + x - 4 πn - π = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 4 πn - π

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

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

化简

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

1^{2} - 4 \cdot 1 \cdot (- 4 πn - π)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 4 πn - π)

数字相乘:

4 \cdot 1 = 4

= 1 - 4 (- 4 πn - π)

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

将解分隔开

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

x = \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2 \cdot 1} : \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2}

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

数字相乘:

2 \cdot 1 = 2

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

x = \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2 \cdot 1} : \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2}

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

数字相乘:

2 \cdot 1 = 2

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

二次方程组的解是:

x = \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2}

解

\frac{x + x ^{2}}{2} = \frac{3 π}{2} + 2 πn : x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}

\frac{x + x ^{2}}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{x + x ^{2}}{2} = \frac{3 π}{2} + 2 πn

在两边乘以

2

\frac{x + x ^{2}}{2} \cdot 2 = \frac{3 π}{2} \cdot 2 + 2 πn \cdot 2

化简

x + x^{2} = 3 π + 4 πn

x + x^{2} = 3 π + 4 πn

将

4 πn

para o lado esquerdo

x + x^{2} = 3 π + 4 πn

两边减去

4 πn

x + x^{2} - 4 πn = 3 π + 4 πn - 4 πn

化简

x + x^{2} - 4 πn = 3 π

x + x^{2} - 4 πn = 3 π

将

3 π

para o lado esquerdo

x + x^{2} - 4 πn = 3 π

两边减去

3 π

x + x^{2} - 4 πn - 3 π = 3 π - 3 π

化简

x + x^{2} - 4 πn - 3 π = 0

x + x^{2} - 4 πn - 3 π = 0

改写成标准形式

a x^{2} + b x + c = 0

x^{2} + x - 4 πn - 3 π = 0

使用求根公式求解

x^{2} + x - 4 πn - 3 π = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 4 πn - 3 π

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 4 πn - 3 π )}{2 \cdot 1}

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 4 πn - 3 π )}{2 \cdot 1}

化简

1^{2} - 4 \cdot 1 \cdot (- 4 πn - 3 π) : 1 - 4 (- 4 πn - 3 π)

1^{2} - 4 \cdot 1 \cdot (- 4 πn - 3 π)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 4 πn - 3 π)

数字相乘:

4 \cdot 1 = 4

= 1 - 4 (- 4 πn - 3 π)

x_{1, 2} = \frac{- 1 \pm 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1}, x_{2} = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1}

x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1} : \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}

\frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + - 4 ( - 4 πn - 3 π ) + 1}{2}

x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1} : \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}

\frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - - 4 ( - 4 πn - 3 π ) + 1}{2}

二次方程组的解是:

x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}

x = \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}

sin (\frac{- x + x ^{2}}{2}) = 0 : x = \frac{1 + 1 + 16 πn}{2}, x = \frac{1 - 1 + 16 πn}{2}, x = \frac{1 + 1 - 4 ( - 4 πn - 2 π )}{2}, x = \frac{1 - 1 - 4 ( - 4 πn - 2 π )}{2}

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

sin (\frac{- x + x ^{2}}{2}) = 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}

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

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

解

\frac{- x + x ^{2}}{2} = 0 + 2 πn : x = \frac{1 + 1 + 16 πn}{2}, x = \frac{1 - 1 + 16 πn}{2}

\frac{- x + x ^{2}}{2} = 0 + 2 πn

在两边乘以

2

\frac{- x + x ^{2}}{2} = 0 + 2 πn

在两边乘以

2

\frac{- x + x ^{2}}{2} \cdot 2 = 0 \cdot 2 + 2 πn \cdot 2

化简

- x + x^{2} = 0 + 4 πn

- x + x^{2} = 0 + 4 πn

- x + x^{2} = 4 πn

将

4 πn

para o lado esquerdo

- x + x^{2} = 4 πn

两边减去

4 πn

- x + x^{2} - 4 πn = 4 πn - 4 πn

化简

- x + x^{2} - 4 πn = 0

- x + x^{2} - 4 πn = 0

改写成标准形式

a x^{2} + b x + c = 0

x^{2} - x - 4 πn = 0

使用求根公式求解

x^{2} - x - 4 πn = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 4 πn

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

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

化简

(- 1)^{2} - 4 \cdot 1 \cdot (- 4 πn) : 1 + 16 πn

(- 1)^{2} - 4 \cdot 1 \cdot (- 4 πn)

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \cdot 4 πn

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 4 πn = 16 πn

4 \cdot 1 \cdot 4 πn

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 16 πn

= 1 + 16 πn

x_{1, 2} = \frac{- ( - 1 ) \pm 1 + 16 πn}{2 \cdot 1}

将解分隔开

x_{1} = \frac{- ( - 1 ) + 1 + 16 πn}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 1 + 16 πn}{2 \cdot 1}

x = \frac{- ( - 1 ) + 1 + 16 πn}{2 \cdot 1} : \frac{1 + 1 + 16 πn}{2}

\frac{- ( - 1 ) + 1 + 16 πn}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 1 + 16 πn}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 16 πn + 1}{2}

x = \frac{- ( - 1 ) - 1 + 16 πn}{2 \cdot 1} : \frac{1 - 1 + 16 πn}{2}

\frac{- ( - 1 ) - 1 + 16 πn}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 1 + 16 πn}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 16 πn + 1}{2}

二次方程组的解是:

x = \frac{1 + 1 + 16 πn}{2}, x = \frac{1 - 1 + 16 πn}{2}

解

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

\frac{- x + x ^{2}}{2} = π + 2 πn

在两边乘以

2

\frac{- x + x ^{2}}{2} = π + 2 πn

在两边乘以

2

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

化简

- x + x^{2} = 2 π + 4 πn

- x + x^{2} = 2 π + 4 πn

将

4 πn

para o lado esquerdo

- x + x^{2} = 2 π + 4 πn

两边减去

4 πn

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

化简

- x + x^{2} - 4 πn = 2 π

- x + x^{2} - 4 πn = 2 π

将

2 π

para o lado esquerdo

- x + x^{2} - 4 πn = 2 π

两边减去

2 π

- x + x^{2} - 4 πn - 2 π = 2 π - 2 π

化简

- x + x^{2} - 4 πn - 2 π = 0

- x + x^{2} - 4 πn - 2 π = 0

改写成标准形式

a x^{2} + b x + c = 0

x^{2} - x - 4 πn - 2 π = 0

使用求根公式求解

x^{2} - x - 4 πn - 2 π = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 4 πn - 2 π

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

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

化简

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

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot (- 4 πn - 2 π) = 4 (- 4 πn - 2 π)

4 \cdot 1 \cdot (- 4 πn - 2 π)

数字相乘:

4 \cdot 1 = 4

= 4 (- 4 πn - 2 π)

= 1 - 4 (- 4 πn - 2 π)

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

将解分隔开

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

x = \frac{- ( - 1 ) + 1 - 4 ( - 4 πn - 2 π )}{2 \cdot 1} : \frac{1 + 1 - 4 ( - 4 πn - 2 π )}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

x = \frac{- ( - 1 ) - 1 - 4 ( - 4 πn - 2 π )}{2 \cdot 1} : \frac{1 - 1 - 4 ( - 4 πn - 2 π )}{2}

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

二次方程组的解是:

x = \frac{1 + 1 - 4 ( - 4 πn - 2 π )}{2}, x = \frac{1 - 1 - 4 ( - 4 πn - 2 π )}{2}

x = \frac{1 + 1 + 16 πn}{2}, x = \frac{1 - 1 + 16 πn}{2}, x = \frac{1 + 1 - 4 ( - 4 πn - 2 π )}{2}, x = \frac{1 - 1 - 4 ( - 4 πn - 2 π )}{2}

合并所有解

x = \frac{- 1 + 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - π )}{2}, x = \frac{- 1 + 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{- 1 - 1 - 4 ( - 4 πn - 3 π )}{2}, x = \frac{1 + 1 + 16 πn}{2}, x = \frac{1 - 1 + 16 πn}{2}, x = \frac{1 + 1 - 4 ( - 4 πn - 2 π )}{2}, x = \frac{1 - 1 - 4 ( - 4 πn - 2 π )}{2}

作图

流行的例子

tan(θ)=(sqrt(2))/2 csc(θ)

tan (θ) = \frac{2}{2} csc (θ)

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

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

tan (θ) = 0.75

cos (6 x) = 1

csc^2(θ)-2csc(θ)=0

csc^{2} (θ) - 2 csc (θ) = 0