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受欢迎的三角函数 >

2cos^2(3x)-3cos(3x)=-1,0<= x<= 2pi

解答

2 cos^{2} (3 x) - 3 cos (3 x) = - 1, 0 \leq x \leq 2 π

解答

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π, x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

+1

度数

x = 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}, x = 36 0^{\circ}, x = 2 0^{\circ}, x = 10 0^{\circ}, x = 14 0^{\circ}, x = 22 0^{\circ}, x = 26 0^{\circ}, x = 34 0^{\circ}

求解步骤

2 cos^{2} (3 x) - 3 cos (3 x) = - 1, 0 \leq x \leq 2 π

用替代法求解

2 cos^{2} (3 x) - 3 cos (3 x) = - 1

令

: cos (3 x) = u

2 u^{2} - 3 u = - 1

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

2 u^{2} - 3 u = - 1

将

1

para o lado esquerdo

2 u^{2} - 3 u = - 1

两边加上

1

2 u^{2} - 3 u + 1 = - 1 + 1

化简

2 u^{2} - 3 u + 1 = 0

2 u^{2} - 3 u + 1 = 0

使用求根公式求解

2 u^{2} - 3 u + 1 = 0

二次方程求根公式:

若

a = 2, b = - 3, c = 1

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

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

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

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

数字相加:

3 + 1 = 4

= \frac{4}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

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

\frac{- ( - 3 ) - 1}{2 \cdot 2}

使用法则

- (- a) = a

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

数字相减:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = cos (3 x)

代回

cos (3 x) = 1, cos (3 x) = \frac{1}{2}

cos (3 x) = 1, cos (3 x) = \frac{1}{2}

cos (3 x) = 1, 0 \leq x \leq 2 π : x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

cos (3 x) = 1, 0 \leq x \leq 2 π

cos (3 x) = 1

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}

3 x = 0 + 2 πn

3 x = 0 + 2 πn

解

3 x = 0 + 2 πn : x = \frac{2 πn}{3}

3 x = 0 + 2 πn

0 + 2 πn = 2 πn

3 x = 2 πn

两边除以

3

3 x = 2 πn

两边除以

3

\frac{3 x}{3} = \frac{2 πn}{3}

化简

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

在

0 \leq x \leq 2 π

范围内的解

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

cos (3 x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

cos (3 x) = \frac{1}{2}, 0 \leq x \leq 2 π

cos (3 x) = \frac{1}{2}

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}

3 x = \frac{π}{3} + 2 πn, 3 x = \frac{5 π}{3} + 2 πn

3 x = \frac{π}{3} + 2 πn, 3 x = \frac{5 π}{3} + 2 πn

解

3 x = \frac{π}{3} + 2 πn : x = \frac{π}{9} + \frac{2 πn}{3}

3 x = \frac{π}{3} + 2 πn

两边除以

3

3 x = \frac{π}{3} + 2 πn

两边除以

3

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

化简

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

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3} : \frac{π}{9} + \frac{2 πn}{3}

\frac{\frac{π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{π}{3}}{3} = \frac{π}{9}

\frac{\frac{π}{3}}{3}

使用分式法则:

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

= \frac{π}{3 \cdot 3}

数字相乘:

3 \cdot 3 = 9

= \frac{π}{9}

= \frac{π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}

解

3 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{9} + \frac{2 πn}{3}

3 x = \frac{5 π}{3} + 2 πn

两边除以

3

3 x = \frac{5 π}{3} + 2 πn

两边除以

3

\frac{3 x}{3} = \frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3}

化简

\frac{3 x}{3} = \frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3} : \frac{5 π}{9} + \frac{2 πn}{3}

\frac{\frac{5 π}{3}}{3} + \frac{2 πn}{3}

\frac{\frac{5 π}{3}}{3} = \frac{5 π}{9}

\frac{\frac{5 π}{3}}{3}

使用分式法则:

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

= \frac{5 π}{3 \cdot 3}

数字相乘:

3 \cdot 3 = 9

= \frac{5 π}{9}

= \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{5 π}{9} + \frac{2 πn}{3}

x = \frac{π}{9} + \frac{2 πn}{3}, x = \frac{5 π}{9} + \frac{2 πn}{3}

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

合并所有解

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π, x = \frac{π}{9}, x = \frac{5 π}{9}, x = \frac{7 π}{9}, x = \frac{11 π}{9}, x = \frac{13 π}{9}, x = \frac{17 π}{9}

作图

流行的例子

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