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受欢迎的三角函数 >

8sin(x)+6cos(x)=8

解答

8 sin (x) + 6 cos (x) = 8

解答

x = \frac{π}{2} + 2 πn, x = 0.28379 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 16.26020 \dots^{\circ} + 36 0^{\circ} n

求解步骤

8 sin (x) + 6 cos (x) = 8

两边减去

6 cos (x)

8 sin (x) = 8 - 6 cos (x)

两边进行平方

(8 sin (x))^{2} = (8 - 6 cos (x))^{2}

两边减去

(8 - 6 cos (x))^{2}

64 sin^{2} (x) - 64 + 96 cos (x) - 36 cos^{2} (x) = 0

使用三角恒等式改写

- 64 - 36 cos^{2} (x) + 64 sin^{2} (x) + 96 cos (x)

使用毕达哥拉斯恒等式:

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

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

= - 64 - 36 cos^{2} (x) + 64 (1 - cos^{2} (x)) + 96 cos (x)

化简

- 64 - 36 cos^{2} (x) + 64 (1 - cos^{2} (x)) + 96 cos (x) : 96 cos (x) - 100 cos^{2} (x)

- 64 - 36 cos^{2} (x) + 64 (1 - cos^{2} (x)) + 96 cos (x)

乘开

64 (1 - cos^{2} (x)) : 64 - 64 cos^{2} (x)

64 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 64, b = 1, c = cos^{2} (x)

= 64 \cdot 1 - 64 cos^{2} (x)

数字相乘:

64 \cdot 1 = 64

= 64 - 64 cos^{2} (x)

= - 64 - 36 cos^{2} (x) + 64 - 64 cos^{2} (x) + 96 cos (x)

化简

- 64 - 36 cos^{2} (x) + 64 - 64 cos^{2} (x) + 96 cos (x) : 96 cos (x) - 100 cos^{2} (x)

- 64 - 36 cos^{2} (x) + 64 - 64 cos^{2} (x) + 96 cos (x)

对同类项分组

= - 36 cos^{2} (x) - 64 cos^{2} (x) + 96 cos (x) - 64 + 64

同类项相加:

- 36 cos^{2} (x) - 64 cos^{2} (x) = - 100 cos^{2} (x)

= - 100 cos^{2} (x) + 96 cos (x) - 64 + 64

- 64 + 64 = 0

= 96 cos (x) - 100 cos^{2} (x)

= 96 cos (x) - 100 cos^{2} (x)

= 96 cos (x) - 100 cos^{2} (x)

- 100 cos^{2} (x) + 96 cos (x) = 0

用替代法求解

- 100 cos^{2} (x) + 96 cos (x) = 0

令

: cos (x) = u

- 100 u^{2} + 96 u = 0

- 100 u^{2} + 96 u = 0 : u = 0, u = \frac{24}{25}

- 100 u^{2} + 96 u = 0

使用求根公式求解

- 100 u^{2} + 96 u = 0

二次方程求根公式:

若

a = - 100, b = 96, c = 0

u_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 ( - 100 ) \cdot 0}{2 ( - 100 )}

u_{1, 2} = \frac{- 96 \pm 9 6 ^{2} - 4 ( - 100 ) \cdot 0}{2 ( - 100 )}

9 6^{2} - 4 (- 100) \cdot 0 = 96

9 6^{2} - 4 (- 100) \cdot 0

使用法则

- (- a) = a

= 9 6^{2} + 4 \cdot 100 \cdot 0

使用法则

0 \cdot a = 0

= 9 6^{2} + 0

9 6^{2} + 0 = 9 6^{2}

= 9 6^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 96

u_{1, 2} = \frac{- 96 \pm 96}{2 ( - 100 )}

将解分隔开

u_{1} = \frac{- 96 + 96}{2 ( - 100 )}, u_{2} = \frac{- 96 - 96}{2 ( - 100 )}

u = \frac{- 96 + 96}{2 ( - 100 )} : 0

\frac{- 96 + 96}{2 ( - 100 )}

去除括号:

(- a) = - a

= \frac{- 96 + 96}{- 2 \cdot 100}

数字相加/相减:

- 96 + 96 = 0

= \frac{0}{- 2 \cdot 100}

数字相乘:

2 \cdot 100 = 200

= \frac{0}{- 200}

使用分式法则:

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

= - \frac{0}{200}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 96 - 96}{2 ( - 100 )} : \frac{24}{25}

\frac{- 96 - 96}{2 ( - 100 )}

去除括号:

(- a) = - a

= \frac{- 96 - 96}{- 2 \cdot 100}

数字相减:

- 96 - 96 = - 192

= \frac{- 192}{- 2 \cdot 100}

数字相乘:

2 \cdot 100 = 200

= \frac{- 192}{- 200}

使用分式法则:

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

= \frac{192}{200}

约分:

8

= \frac{24}{25}

二次方程组的解是:

u = 0, u = \frac{24}{25}

u = cos (x)

代回

cos (x) = 0, cos (x) = \frac{24}{25}

cos (x) = 0, cos (x) = \frac{24}{25}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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}

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

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

cos (x) = \frac{24}{25} : x = arccos (\frac{24}{25}) + 2 πn, x = 2 π - arccos (\frac{24}{25}) + 2 πn

cos (x) = \frac{24}{25}

使用反三角函数性质

cos (x) = \frac{24}{25}

cos (x) = \frac{24}{25}

的通解

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

x = arccos (\frac{24}{25}) + 2 πn, x = 2 π - arccos (\frac{24}{25}) + 2 πn

x = arccos (\frac{24}{25}) + 2 πn, x = 2 π - arccos (\frac{24}{25}) + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (\frac{24}{25}) + 2 πn, x = 2 π - arccos (\frac{24}{25}) + 2 πn

将解代入原方程进行验证

将它们代入

8 sin (x) + 6 cos (x) = 8

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

8 sin (x) + 6 cos (x) = 8

代入

x = \frac{π}{2} + 2 π 1

8 sin (\frac{π}{2} + 2 π 1) + 6 cos (\frac{π}{2} + 2 π 1) = 8

整理后得

8 = 8

\Rightarrow 真

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

8 sin (x) + 6 cos (x) = 8

代入

x = \frac{3 π}{2} + 2 π 1

8 sin (\frac{3 π}{2} + 2 π 1) + 6 cos (\frac{3 π}{2} + 2 π 1) = 8

整理后得

- 8 = 8

\Rightarrow 假

检验

arccos (\frac{24}{25}) + 2 πn

的解:

真

arccos (\frac{24}{25}) + 2 πn

代入

n = 1

arccos (\frac{24}{25}) + 2 π 1

对于

8 sin (x) + 6 cos (x) = 8

代入

x = arccos (\frac{24}{25}) + 2 π 1

8 sin (arccos (\frac{24}{25}) + 2 π 1) + 6 cos (arccos (\frac{24}{25}) + 2 π 1) = 8

整理后得

8 = 8

\Rightarrow 真

检验

2 π - arccos (\frac{24}{25}) + 2 πn

的解:

假

2 π - arccos (\frac{24}{25}) + 2 πn

代入

n = 1

2 π - arccos (\frac{24}{25}) + 2 π 1

对于

8 sin (x) + 6 cos (x) = 8

代入

x = 2 π - arccos (\frac{24}{25}) + 2 π 1

8 sin (2 π - arccos (\frac{24}{25}) + 2 π 1) + 6 cos (2 π - arccos (\frac{24}{25}) + 2 π 1) = 8

整理后得

3.52 = 8

\Rightarrow 假

x = \frac{π}{2} + 2 πn, x = arccos (\frac{24}{25}) + 2 πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = 0.28379 \dots + 2 πn

作图

流行的例子

2tan^2(x)-5tan(x)+3=0

2 tan^{2} (x) - 5 tan (x) + 3 = 0

sin (x) = \frac{2}{9}

sin (x) = \frac{2}{4}

3 - tan^{2} (x) = 0

csc (θ) = - \frac{8}{5}