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受欢迎的三角函数 >

sqrt(3)sin(x)-2cos(x)=sqrt(7)

解答

3 sin (x) - 2 cos (x) = 7

解答

x = 2.42786 \dots + 2 πn

+1

度数

x = 139.10660 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 sin (x) - 2 cos (x) = 7

两边加上

2 cos (x)

3 sin (x) = 7 + 2 cos (x)

两边进行平方

(3 sin (x))^{2} = (7 + 2 cos (x))^{2}

两边减去

(7 + 2 cos (x))^{2}

3 sin^{2} (x) - 7 - 47 cos (x) - 4 cos^{2} (x) = 0

使用三角恒等式改写

- 7 + 3 sin^{2} (x) - 4 cos^{2} (x) - 4 cos (x) 7

使用毕达哥拉斯恒等式:

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

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

= - 7 + 3 (1 - cos^{2} (x)) - 4 cos^{2} (x) - 4 cos (x) 7

化简

- 7 + 3 (1 - cos^{2} (x)) - 4 cos^{2} (x) - 4 cos (x) 7 : - 7 cos^{2} (x) - 47 cos (x) - 4

- 7 + 3 (1 - cos^{2} (x)) - 4 cos^{2} (x) - 4 cos (x) 7

= - 7 + 3 (1 - cos^{2} (x)) - 4 cos^{2} (x) - 47 cos (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = cos^{2} (x)

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

数字相乘:

3 \cdot 1 = 3

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

= - 7 + 3 - 3 cos^{2} (x) - 4 cos^{2} (x) - 4 cos (x) 7

化简

- 7 + 3 - 3 cos^{2} (x) - 4 cos^{2} (x) - 4 cos (x) 7 : - 7 cos^{2} (x) - 47 cos (x) - 4

- 7 + 3 - 3 cos^{2} (x) - 4 cos^{2} (x) - 4 cos (x) 7

同类项相加:

- 3 cos^{2} (x) - 4 cos^{2} (x) = - 7 cos^{2} (x)

= - 7 + 3 - 7 cos^{2} (x) - 47 cos (x)

数字相加/相减:

- 7 + 3 = - 4

= - 7 cos^{2} (x) - 47 cos (x) - 4

= - 7 cos^{2} (x) - 47 cos (x) - 4

= - 7 cos^{2} (x) - 47 cos (x) - 4

- 4 - 7 cos^{2} (x) - 4 cos (x) 7 = 0

用替代法求解

- 4 - 7 cos^{2} (x) - 4 cos (x) 7 = 0

令

: cos (x) = u

- 4 - 7 u^{2} - 4 u 7 = 0

- 4 - 7 u^{2} - 4 u 7 = 0 : u = - \frac{2 7}{7}

- 4 - 7 u^{2} - 4 u 7 = 0

改写成标准形式

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

- 7 u^{2} - 47 u - 4 = 0

使用求根公式求解

- 7 u^{2} - 47 u - 4 = 0

二次方程求根公式:

若

a = - 7, b = - 47, c = - 4

u_{1, 2} = \frac{- ( - 4 7 ) \pm ( - 4 7 ) ^{2} - 4 ( - 7 ) ( - 4 )}{2 ( - 7 )}

u_{1, 2} = \frac{- ( - 4 7 ) \pm ( - 4 7 ) ^{2} - 4 ( - 7 ) ( - 4 )}{2 ( - 7 )}

(- 47)^{2} - 4 (- 7) (- 4) = 0

(- 47)^{2} - 4 (- 7) (- 4)

使用法则

- (- a) = a

= (- 47)^{2} - 4 \cdot 7 \cdot 4

(- 47)^{2} = 4^{2} \cdot 7

(- 47)^{2}

使用指数法则:

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

若

n

是偶数

(- 47)^{2} = (47)^{2}

= (47)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 4^{2} (7)^{2}

(7)^{2} : 7

使用根式运算法则:

a = a^{\frac{1}{2}}

= (7^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 7^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 7

= 4^{2} \cdot 7

4 \cdot 7 \cdot 4 = 112

4 \cdot 7 \cdot 4

数字相乘:

4 \cdot 7 \cdot 4 = 112

= 112

= 4^{2} \cdot 7 - 112

4^{2} \cdot 7 = 112

4^{2} \cdot 7

4^{2} = 16

= 16 \cdot 7

数字相乘:

16 \cdot 7 = 112

= 112

= 112 - 112

数字相减:

112 - 112 = 0

= 0

u_{1, 2} = \frac{- ( - 4 7 ) \pm 0}{2 ( - 7 )}

u = \frac{- ( - 4 7 )}{2 ( - 7 )}

\frac{- ( - 4 7 )}{2 ( - 7 )} = - \frac{2 7}{7}

\frac{- ( - 4 7 )}{2 ( - 7 )}

去除括号:

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

= \frac{4 7}{- 2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{4 7}{- 14}

使用分式法则:

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

= - \frac{4 7}{14}

约分:

2

= - \frac{2 7}{7}

u = - \frac{2 7}{7}

二次方程组的解是:

u = - \frac{2 7}{7}

u = cos (x)

代回

cos (x) = - \frac{2 7}{7}

cos (x) = - \frac{2 7}{7}

cos (x) = - \frac{2 7}{7} : x = arccos (- \frac{2 7}{7}) + 2 πn, x = - arccos (- \frac{2 7}{7}) + 2 πn

cos (x) = - \frac{2 7}{7}

使用反三角函数性质

cos (x) = - \frac{2 7}{7}

cos (x) = - \frac{2 7}{7}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (- \frac{2 7}{7}) + 2 πn, x = - arccos (- \frac{2 7}{7}) + 2 πn

x = arccos (- \frac{2 7}{7}) + 2 πn, x = - arccos (- \frac{2 7}{7}) + 2 πn

合并所有解

x = arccos (- \frac{2 7}{7}) + 2 πn, x = - arccos (- \frac{2 7}{7}) + 2 πn

将解代入原方程进行验证

将它们代入

3 sin (x) - 2 cos (x) = 7

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{2 7}{7}) + 2 πn

的解:

真

arccos (- \frac{2 7}{7}) + 2 πn

代入

n = 1

arccos (- \frac{2 7}{7}) + 2 π 1

对于

3 sin (x) - 2 cos (x) = 7

代入

x = arccos (- \frac{2 7}{7}) + 2 π 1

3 sin (arccos (- \frac{2 7}{7}) + 2 π 1) - 2 cos (arccos (- \frac{2 7}{7}) + 2 π 1) = 7

整理后得

2.64575 \dots = 2.64575 \dots

\Rightarrow 真

检验

- arccos (- \frac{2 7}{7}) + 2 πn

的解:

假

- arccos (- \frac{2 7}{7}) + 2 πn

代入

n = 1

- arccos (- \frac{2 7}{7}) + 2 π 1

对于

3 sin (x) - 2 cos (x) = 7

代入

x = - arccos (- \frac{2 7}{7}) + 2 π 1

3 sin (- arccos (- \frac{2 7}{7}) + 2 π 1) - 2 cos (- arccos (- \frac{2 7}{7}) + 2 π 1) = 7

整理后得

0.37796 \dots = 2.64575 \dots

\Rightarrow 假

x = arccos (- \frac{2 7}{7}) + 2 πn

以小数形式表示解

x = 2.42786 \dots + 2 πn

作图

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