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受欢迎的三角函数 >

2sin^3(x)-5sin^2(x)+2sin(x)=0

解答

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

解答

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

用替代法求解

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

令

: sin (x) = u

2 u^{3} - 5 u^{2} + 2 u = 0

2 u^{3} - 5 u^{2} + 2 u = 0 : u = 0, u = \frac{1}{2}, u = 2

2 u^{3} - 5 u^{2} + 2 u = 0

因式分解

2 u^{3} - 5 u^{2} + 2 u : u (2 u - 1) (u - 2)

2 u^{3} - 5 u^{2} + 2 u

因式分解出通项

u : u (2 u^{2} - 5 u + 2)

2 u^{3} - 5 u^{2} + 2 u

使用指数法则:

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

u^{2} = uu

= 2 u^{2} u - 5 uu + 2 u

因式分解出通项

u

= u (2 u^{2} - 5 u + 2)

= u (2 u^{2} - 5 u + 2)

分解

2 u^{2} - 5 u + 2 : (2 u - 1) (u - 2)

2 u^{2} - 5 u + 2

将表达式拆分成组

2 u^{2} - 5 u + 2

定义

4

的因数:

1, 2, 4

4

约数 (因数)

找到

4

的质因数:

2, 2

4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2

添加质因数:

2

将 1 和数字

4

自身相加

1, 4

4

的因数

1, 2, 4

4

的负因数:

- 1, - 2, - 4

将因数乘以

- 1

得到负因数

- 1, - 2, - 4

对于每两个因数

u * v = 4

，检验是否

u + v = - 5

检验

u = 1, v = 4 : u * v = 4, u + v = 5 \Rightarrow

假检验

u = 2, v = 2 : u * v = 4, u + v = 4 \Rightarrow

假

u = - 1, v = - 4

分组为

(a x^{2} + ux) + (vx + c)

(2 u^{2} - u) + (- 4 u + 2)

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

从

2 u^{2} - u

分解出因式

u

:

u (2 u - 1)

2 u^{2} - u

使用指数法则:

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

u^{2} = uu

= 2 uu - u

因式分解出通项

u

= u (2 u - 1)

从

- 4 u + 2

分解出因式

- 2

:

- 2 (2 u - 1)

- 4 u + 2

将

4

改写为

2 \cdot 2

= - 2 \cdot 2 u + 2

因式分解出通项

- 2

= - 2 (2 u - 1)

= u (2 u - 1) - 2 (2 u - 1)

因式分解出通项

2 u - 1

= (2 u - 1) (u - 2)

= u (2 u - 1) (u - 2)

u (2 u - 1) (u - 2) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or 2 u - 1 = 0 or u - 2 = 0

解

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

将

1

到右边

2 u - 1 = 0

两边加上

1

2 u - 1 + 1 = 0 + 1

化简

2 u = 1

2 u = 1

两边除以

2

2 u = 1

两边除以

2

\frac{2 u}{2} = \frac{1}{2}

化简

u = \frac{1}{2}

u = \frac{1}{2}

解

u - 2 = 0 : u = 2

u - 2 = 0

将

2

到右边

u - 2 = 0

两边加上

2

u - 2 + 2 = 0 + 2

化简

u = 2

u = 2

解为

u = 0, u = \frac{1}{2}, u = 2

u = sin (x)

代回

sin (x) = 0, sin (x) = \frac{1}{2}, sin (x) = 2

sin (x) = 0, sin (x) = \frac{1}{2}, sin (x) = 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} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

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

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = 2 :

无解

sin (x) = 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

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