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受欢迎的三角函数 >

sin(x)=sin(x^2)

解答

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

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}

+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

求解步骤

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

两边减去

sin (x^{2})

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

无解

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 ( - 1 ) ( - 4 πn )}{2 ( - 1 )}

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

化简

1^{2} - 4 (- 1) (- 4 πn) : 1 - 16 πn

1^{2} - 4 (- 1) (- 4 πn)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 4 πn)

使用法则

- (- a) = a

= 1 - 4 \cdot 1 \cdot 4 πn

数字相乘:

4 \cdot 1 \cdot 4 = 16

= 1 - 16 πn

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

将解分隔开

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

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

\frac{- 1 + 1 - 16 πn}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 1 - 16 πn}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + - 16 πn + 1}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 1 - 16 πn}{2}

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

\frac{- 1 - 1 - 16 πn}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 - 1 - 16 πn}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - - 16 πn + 1}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

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

化简

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

1^{2} - 4 (- 1) (- 4 πn - 2 π)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 4 πn - 2 π)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot (- 4 πn - 2 π)

数字相乘:

4 \cdot 1 = 4

= 1 + 4 (- 4 πn - 2 π)

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

将解分隔开

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

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

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

去除括号:

(- 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}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

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

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

去除括号:

(- 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}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

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

因为方程对以下值无定义:

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

无解

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}

作图

流行的例子

sin (x) = - 0.23

2sin^2(x)+3sin(x)+1=0,0<= x<= 2pi

2 sin^{2} (x) + 3 sin (x) + 1 = 0, 0 \leq x \leq 2 π

cos(3x)-1=sin(3x)

cos (3 x) - 1 = sin (3 x)

sin(x/3)=(sqrt(2))/2

sin (\frac{x}{3}) = \frac{2}{2}

2cos(3x)=-sqrt(3)

2 cos (3 x) = - 3