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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

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

求解步骤

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

两边减去

2

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

令

: u = \frac{x}{2}

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

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

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

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 2 + 1 = - 1

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

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

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

- 1 - 2 cos^{2} (u) + 3 cos (u) = 0

用替代法求解

- 1 - 2 cos^{2} (u) + 3 cos (u) = 0

令

: cos (u) = u

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

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

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

改写成标准形式

a x^{2} + b x + c = 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 ( - 2 ) ( - 1 )}{2 ( - 2 )}

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

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

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

使用法则

- (- a) = a

= 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 ( - 2 )}

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/相减:

- 3 + 1 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

去除括号:

(- a) = - a

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

数字相减:

- 3 - 1 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

使用分式法则:

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

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

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

u = cos (u)

代回

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

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

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

cos (u) = 1 : u = 2 πn

cos (u) = 1

cos (u) = 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}

u = 0 + 2 πn

u = 0 + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn

合并所有解

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

\frac{x}{2} = 2 πn : x = 4 πn

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{x}{2} = 2 πn

在两边乘以

2

\frac{2 x}{2} = 2 \cdot 2 πn

化简

x = 4 πn

x = 4 πn

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

作图

流行的例子

tan(x)-csc(2x)=0

tan (x) - csc (2 x) = 0

solvefor x,z=sin(x^2+y^2)

so l v e f or x, z = sin (x^{2} + y^{2})

tan(θ)=(78.48)/(196.2)

tan (θ) = \frac{78.48}{196.2}

sec^2(x)-tan(x)=1,0<= x<2pi

sec^{2} (x) - tan (x) = 1, 0 \leq x < 2 π

cot(θ)+1=0,(0,2pi)

cot (θ) + 1 = 0, (0, 2 π)