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受欢迎的三角函数 >

3+2cos(x)=4cos(x/2)

解答

3 + 2 cos (x) = 4 cos (\frac{x}{2})

解答

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

+1

度数

x = 12 0^{\circ} + 72 0^{\circ} n, x = 60 0^{\circ} + 72 0^{\circ} n

求解步骤

3 + 2 cos (x) = 4 cos (\frac{x}{2})

两边减去

4 cos (\frac{x}{2})

3 + 2 cos (x) - 4 cos (\frac{x}{2}) = 0

令

: u = \frac{x}{2}

3 + 2 cos (2 u) - 4 cos (u) = 0

使用三角恒等式改写

3 + 2 cos (2 u) - 4 cos (u)

使用倍角公式:

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

= 3 + 2 (2 cos^{2} (u) - 1) - 4 cos (u)

化简

3 + 2 (2 cos^{2} (u) - 1) - 4 cos (u) : 4 cos^{2} (u) - 4 cos (u) + 1

3 + 2 (2 cos^{2} (u) - 1) - 4 cos (u)

乘开

2 (2 cos^{2} (u) - 1) : 4 cos^{2} (u) - 2

2 (2 cos^{2} (u) - 1)

使用分配律:

a (b - c) = ab - a c

a = 2, b = 2 cos^{2} (u), c = 1

= 2 \cdot 2 cos^{2} (u) - 2 \cdot 1

化简

2 \cdot 2 cos^{2} (u) - 2 \cdot 1 : 4 cos^{2} (u) - 2

2 \cdot 2 cos^{2} (u) - 2 \cdot 1

数字相乘:

2 \cdot 2 = 4

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

数字相乘:

2 \cdot 1 = 2

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

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

= 3 + 4 cos^{2} (u) - 2 - 4 cos (u)

化简

3 + 4 cos^{2} (u) - 2 - 4 cos (u) : 4 cos^{2} (u) - 4 cos (u) + 1

3 + 4 cos^{2} (u) - 2 - 4 cos (u)

对同类项分组

= 4 cos^{2} (u) - 4 cos (u) + 3 - 2

数字相加/相减:

3 - 2 = 1

= 4 cos^{2} (u) - 4 cos (u) + 1

= 4 cos^{2} (u) - 4 cos (u) + 1

= 4 cos^{2} (u) - 4 cos (u) + 1

1 - 4 cos (u) + 4 cos^{2} (u) = 0

用替代法求解

1 - 4 cos (u) + 4 cos^{2} (u) = 0

令

: cos (u) = u

1 - 4 u + 4 u^{2} = 0

1 - 4 u + 4 u^{2} = 0 : u = \frac{1}{2}

1 - 4 u + 4 u^{2} = 0

改写成标准形式

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

4 u^{2} - 4 u + 1 = 0

使用求根公式求解

4 u^{2} - 4 u + 1 = 0

二次方程求根公式:

若

a = 4, b = - 4, c = 1

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

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

(- 4)^{2} - 4 \cdot 4 \cdot 1 = 0

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

u = \frac{- ( - 4 )}{2 \cdot 4}

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

\frac{- ( - 4 )}{2 \cdot 4}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{8}

约分:

4

= \frac{1}{2}

u = \frac{1}{2}

二次方程组的解是:

u = \frac{1}{2}

u = cos (u)

代回

cos (u) = \frac{1}{2}

cos (u) = \frac{1}{2}

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

cos (u) = \frac{1}{2}

cos (u) = \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}

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

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

合并所有解

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

u = \frac{x}{2}

代回

\frac{x}{2} = \frac{π}{3} + 2 πn : x = \frac{2 π}{3} + 4 πn

\frac{x}{2} = \frac{π}{3} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{π}{3} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{π}{3} + 2 \cdot 2 πn : \frac{2 π}{3} + 4 πn

2 \cdot \frac{π}{3} + 2 \cdot 2 πn

乘

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

分式相乘:

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

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= \frac{2 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{3} + 2 πn : x = \frac{10 π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{3} + 2 πn

在两边乘以

2

\frac{x}{2} = \frac{5 π}{3} + 2 πn

在两边乘以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{3} = \frac{10 π}{3}

2 \cdot \frac{5 π}{3}

分式相乘:

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

= \frac{5 π 2}{3}

数字相乘:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数字相乘:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

x = \frac{10 π}{3} + 4 πn

x = \frac{10 π}{3} + 4 πn

x = \frac{10 π}{3} + 4 πn

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

作图

流行的例子

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

- 3 sin (x) = cos (x)

2csc^2(x)=5-5cot(x)

2 csc^{2} (x) = 5 - 5 cot (x)

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

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

5 + 2 sin (x) - 7 = 0

0.5 = 1 - sin (θ)