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受欢迎的三角函数 >

cos(4x)+1=3sin(2x)

解答

cos (4 x) + 1 = 3 sin (2 x)

解答

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

+1

度数

x = 1 5^{\circ} + 18 0^{\circ} n, x = 7 5^{\circ} + 18 0^{\circ} n

求解步骤

cos (4 x) + 1 = 3 sin (2 x)

两边减去

3 sin (2 x)

cos (4 x) + 1 - 3 sin (2 x) = 0

使用三角恒等式改写

1 + cos (4 x) - 3 sin (2 x)

cos (4 x) = 2 cos^{2} (2 x) - 1

cos (4 x)

改写为

= cos (2 \cdot 2 x)

使用倍角公式:

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

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

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

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

化简

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

1 + 2 cos^{2} (2 x) - 1 - 3 sin (2 x)

对同类项分组

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

1 - 1 = 0

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

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

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

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

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

(1 - sin^{2} (2 x)) \cdot 2 - 3 sin (2 x) = 0

用替代法求解

(1 - sin^{2} (2 x)) \cdot 2 - 3 sin (2 x) = 0

令

: sin (2 x) = u

(1 - u^{2}) \cdot 2 - 3 u = 0

(1 - u^{2}) \cdot 2 - 3 u = 0 : u = - 2, u = \frac{1}{2}

(1 - u^{2}) \cdot 2 - 3 u = 0

展开

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

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

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

乘开

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

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

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

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

2 - 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 + 2 = 0

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

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

使用法则

- (- a) = a

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

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 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 + 5}{- 2 \cdot 2}

数字相加:

3 + 5 = 8

= \frac{8}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{- 4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 - 5}{- 2 \cdot 2}

数字相减:

3 - 5 = - 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}

二次方程组的解是:

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

u = sin (2 x)

代回

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

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

sin (2 x) = - 2 :

无解

sin (2 x) = - 2

- 1 \leq sin (x) \leq 1

无解

sin (2 x) = \frac{1}{2} : x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

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

sin (2 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}

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

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

解

2 x = \frac{π}{6} + 2 πn : x = \frac{π}{12} + πn

2 x = \frac{π}{6} + 2 πn

两边除以

2

2 x = \frac{π}{6} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2} : \frac{π}{12} + πn

\frac{\frac{π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

使用分式法则:

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

= \frac{π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

x = \frac{π}{12} + πn

解

2 x = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{12} + πn

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

2 x = \frac{5 π}{6} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

合并所有解

x = \frac{π}{12} + πn, x = \frac{5 π}{12} + πn

作图

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