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受欢迎的三角函数 >

sin(x)+sin^2(x)+sin^3(x)=0

解答

sin (x) + sin^{2} (x) + sin^{3} (x) = 0

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) + sin^{2} (x) + sin^{3} (x) = 0

用替代法求解

sin (x) + sin^{2} (x) + sin^{3} (x) = 0

令

: sin (x) = u

u + u^{2} + u^{3} = 0

u + u^{2} + u^{3} = 0 : u = 0, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u + u^{2} + u^{3} = 0

因式分解

u + u^{2} + u^{3} : u (u^{2} + u + 1)

u + u^{2} + u^{3}

使用指数法则:

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

u^{2} = uu

= u^{2} u + uu + u

因式分解出通项

u

= u (u^{2} + u + 1)

u (u^{2} + u + 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or u^{2} + u + 1 = 0

解

u^{2} + u + 1 = 0 : u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u^{2} + u + 1 = 0

使用求根公式求解

u^{2} + u + 1 = 0

二次方程求根公式:

若

a = 1, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

化简

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

1^{2} - 4 \cdot 1 \cdot 1

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

使用根式运算法则:

- a = - 1 a

- 3 = - 1 3

= - 1 3

使用虚数运算法则:

- 1 = i

= 3 i

u_{1, 2} = \frac{- 1 \pm 3 i}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 3 i}{2 \cdot 1}, u_{2} = \frac{- 1 - 3 i}{2 \cdot 1}

u = \frac{- 1 + 3 i}{2 \cdot 1} : - \frac{1}{2} + i \frac{3}{2}

\frac{- 1 + 3 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 3 i}{2}

将

\frac{- 1 + 3 i}{2}

改写成标准复数形式:

- \frac{1}{2} + \frac{3}{2} i

\frac{- 1 + 3 i}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 3 i}{2} = - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3 i}{2}

= - \frac{1}{2} + \frac{3}{2} i

u = \frac{- 1 - 3 i}{2 \cdot 1} : - \frac{1}{2} - i \frac{3}{2}

\frac{- 1 - 3 i}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 3 i}{2}

将

\frac{- 1 - 3 i}{2}

改写成标准复数形式:

- \frac{1}{2} - \frac{3}{2} i

\frac{- 1 - 3 i}{2}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 3 i}{2} = - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3 i}{2}

= - \frac{1}{2} - \frac{3}{2} i

二次方程组的解是:

u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

解为

u = 0, u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u = sin (x)

代回

sin (x) = 0, sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

sin (x) = 0, sin (x) = - \frac{1}{2} + i \frac{3}{2}, sin (x) = - \frac{1}{2} - i \frac{3}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = - \frac{1}{2} + i \frac{3}{2} :

无解

sin (x) = - \frac{1}{2} + i \frac{3}{2}

无解

sin (x) = - \frac{1}{2} - i \frac{3}{2} :

无解

sin (x) = - \frac{1}{2} - i \frac{3}{2}

无解

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

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