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受欢迎的三角函数 >

2cos(x)-2sqrt(3)*sin(x)=sqrt(8)

解答

2 cos (x) - 23 \cdot sin (x) = 8

解答

x = π + 1.30899 \dots + 2 πn, x = - 0.26179 \dots + 2 πn

+1

度数

x = 25 5^{\circ} + 36 0^{\circ} n, x = - 1 5^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (x) - 23 sin (x) = 8

两边加上

23 sin (x)

2 cos (x) = 22 + 23 sin (x)

两边进行平方

(2 cos (x))^{2} = (22 + 23 sin (x))^{2}

两边减去

(22 + 23 sin (x))^{2}

4 cos^{2} (x) - 8 - 86 sin (x) - 12 sin^{2} (x) = 0

使用三角恒等式改写

- 8 - 12 sin^{2} (x) + 4 cos^{2} (x) - 8 sin (x) 6

使用毕达哥拉斯恒等式:

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

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

= - 8 - 12 sin^{2} (x) + 4 (1 - sin^{2} (x)) - 8 sin (x) 6

化简

- 8 - 12 sin^{2} (x) + 4 (1 - sin^{2} (x)) - 8 sin (x) 6 : - 16 sin^{2} (x) - 86 sin (x) - 4

- 8 - 12 sin^{2} (x) + 4 (1 - sin^{2} (x)) - 8 sin (x) 6

= - 8 - 12 sin^{2} (x) + 4 (1 - sin^{2} (x)) - 86 sin (x)

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

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

= - 8 - 12 sin^{2} (x) + 4 - 4 sin^{2} (x) - 8 sin (x) 6

化简

- 8 - 12 sin^{2} (x) + 4 - 4 sin^{2} (x) - 8 sin (x) 6 : - 16 sin^{2} (x) - 86 sin (x) - 4

- 8 - 12 sin^{2} (x) + 4 - 4 sin^{2} (x) - 8 sin (x) 6

对同类项分组

= - 12 sin^{2} (x) - 4 sin^{2} (x) - 86 sin (x) - 8 + 4

同类项相加:

- 12 sin^{2} (x) - 4 sin^{2} (x) = - 16 sin^{2} (x)

= - 16 sin^{2} (x) - 86 sin (x) - 8 + 4

数字相加/相减:

- 8 + 4 = - 4

= - 16 sin^{2} (x) - 86 sin (x) - 4

= - 16 sin^{2} (x) - 86 sin (x) - 4

= - 16 sin^{2} (x) - 86 sin (x) - 4

- 4 - 16 sin^{2} (x) - 8 sin (x) 6 = 0

用替代法求解

- 4 - 16 sin^{2} (x) - 8 sin (x) 6 = 0

令

: sin (x) = u

- 4 - 16 u^{2} - 8 u 6 = 0

- 4 - 16 u^{2} - 8 u 6 = 0 : u = - \frac{6 + 2}{4}, u = - \frac{6 - 2}{4}

- 4 - 16 u^{2} - 8 u 6 = 0

改写成标准形式

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

- 16 u^{2} - 86 u - 4 = 0

使用求根公式求解

- 16 u^{2} - 86 u - 4 = 0

二次方程求根公式:

若

a = - 16, b = - 86, c = - 4

u_{1, 2} = \frac{- ( - 8 6 ) \pm ( - 8 6 ) ^{2} - 4 ( - 16 ) ( - 4 )}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 6 ) \pm ( - 8 6 ) ^{2} - 4 ( - 16 ) ( - 4 )}{2 ( - 16 )}

(- 86)^{2} - 4 (- 16) (- 4) = 82

(- 86)^{2} - 4 (- 16) (- 4)

使用法则

- (- a) = a

= (- 86)^{2} - 4 \cdot 16 \cdot 4

(- 86)^{2} = 8^{2} \cdot 6

(- 86)^{2}

使用指数法则:

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

若

n

是偶数

(- 86)^{2} = (86)^{2}

= (86)^{2}

使用指数法则:

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

= 8^{2} (6)^{2}

(6)^{2} : 6

使用根式运算法则:

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

= (6^{\frac{1}{2}})^{2}

使用指数法则:

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

= 6^{\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

= 6

= 8^{2} \cdot 6

4 \cdot 16 \cdot 4 = 256

4 \cdot 16 \cdot 4

数字相乘:

4 \cdot 16 \cdot 4 = 256

= 256

= 8^{2} \cdot 6 - 256

8^{2} \cdot 6 = 384

8^{2} \cdot 6

8^{2} = 64

= 64 \cdot 6

数字相乘:

64 \cdot 6 = 384

= 384

= 384 - 256

数字相减:

384 - 256 = 128

= 128

128

质因数分解:

2^{7}

128

128

除以

2128 = 64 \cdot 2

= 2 \cdot 64

64

除以

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 32

32

除以

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{7}

= 2^{7}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{6} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{6}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{6} = 2^{\frac{6}{2}} = 2^{3}

= 2^{3} 2

整理后得

= 82

u_{1, 2} = \frac{- ( - 8 6 ) \pm 8 2}{2 ( - 16 )}

将解分隔开

u_{1} = \frac{- ( - 8 6 ) + 8 2}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 6 ) - 8 2}{2 ( - 16 )}

u = \frac{- ( - 8 6 ) + 8 2}{2 ( - 16 )} : - \frac{6 + 2}{4}

\frac{- ( - 8 6 ) + 8 2}{2 ( - 16 )}

去除括号:

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

= \frac{8 6 + 8 2}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{8 6 + 8 2}{- 32}

使用分式法则:

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

= - \frac{8 6 + 8 2}{32}

消掉

\frac{8 6 + 8 2}{32} : \frac{6 + 2}{4}

\frac{8 6 + 8 2}{32}

因式分解出通项

8

= \frac{8 ( 6 + 2 )}{32}

约分:

8

= \frac{6 + 2}{4}

= - \frac{6 + 2}{4}

u = \frac{- ( - 8 6 ) - 8 2}{2 ( - 16 )} : - \frac{6 - 2}{4}

\frac{- ( - 8 6 ) - 8 2}{2 ( - 16 )}

去除括号:

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

= \frac{8 6 - 8 2}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{8 6 - 8 2}{- 32}

使用分式法则:

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

= - \frac{8 6 - 8 2}{32}

消掉

\frac{8 6 - 8 2}{32} : \frac{6 - 2}{4}

\frac{8 6 - 8 2}{32}

因式分解出通项

8

= \frac{8 ( 6 - 2 )}{32}

约分:

8

= \frac{6 - 2}{4}

= - \frac{6 - 2}{4}

二次方程组的解是:

u = - \frac{6 + 2}{4}, u = - \frac{6 - 2}{4}

u = sin (x)

代回

sin (x) = - \frac{6 + 2}{4}, sin (x) = - \frac{6 - 2}{4}

sin (x) = - \frac{6 + 2}{4}, sin (x) = - \frac{6 - 2}{4}

sin (x) = - \frac{6 + 2}{4} : x = arcsin (- \frac{6 + 2}{4}) + 2 πn, x = π + arcsin (\frac{6 + 2}{4}) + 2 πn

sin (x) = - \frac{6 + 2}{4}

使用反三角函数性质

sin (x) = - \frac{6 + 2}{4}

sin (x) = - \frac{6 + 2}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{6 + 2}{4}) + 2 πn, x = π + arcsin (\frac{6 + 2}{4}) + 2 πn

x = arcsin (- \frac{6 + 2}{4}) + 2 πn, x = π + arcsin (\frac{6 + 2}{4}) + 2 πn

sin (x) = - \frac{6 - 2}{4} : x = arcsin (- \frac{6 - 2}{4}) + 2 πn, x = π + arcsin (\frac{6 - 2}{4}) + 2 πn

sin (x) = - \frac{6 - 2}{4}

使用反三角函数性质

sin (x) = - \frac{6 - 2}{4}

sin (x) = - \frac{6 - 2}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{6 - 2}{4}) + 2 πn, x = π + arcsin (\frac{6 - 2}{4}) + 2 πn

x = arcsin (- \frac{6 - 2}{4}) + 2 πn, x = π + arcsin (\frac{6 - 2}{4}) + 2 πn

合并所有解

x = arcsin (- \frac{6 + 2}{4}) + 2 πn, x = π + arcsin (\frac{6 + 2}{4}) + 2 πn, x = arcsin (- \frac{6 - 2}{4}) + 2 πn, x = π + arcsin (\frac{6 - 2}{4}) + 2 πn

将解代入原方程进行验证

将它们代入

2 cos (x) - 23 sin (x) = 8

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{6 + 2}{4}) + 2 πn

的解:

假

arcsin (- \frac{6 + 2}{4}) + 2 πn

代入

n = 1

arcsin (- \frac{6 + 2}{4}) + 2 π 1

对于

2 cos (x) - 23 sin (x) = 8

代入

x = arcsin (- \frac{6 + 2}{4}) + 2 π 1

2 cos (arcsin (- \frac{6 + 2}{4}) + 2 π 1) - 23 sin (arcsin (- \frac{6 + 2}{4}) + 2 π 1) = 8

整理后得

3.86370 \dots = 2.82842 \dots

\Rightarrow 假

检验

π + arcsin (\frac{6 + 2}{4}) + 2 πn

的解:

真

π + arcsin (\frac{6 + 2}{4}) + 2 πn

代入

n = 1

π + arcsin (\frac{6 + 2}{4}) + 2 π 1

对于

2 cos (x) - 23 sin (x) = 8

代入

x = π + arcsin (\frac{6 + 2}{4}) + 2 π 1

2 cos (π + arcsin (\frac{6 + 2}{4}) + 2 π 1) - 23 sin (π + arcsin (\frac{6 + 2}{4}) + 2 π 1) = 8

整理后得

2.82842 \dots = 2.82842 \dots

\Rightarrow 真

检验

arcsin (- \frac{6 - 2}{4}) + 2 πn

的解:

真

arcsin (- \frac{6 - 2}{4}) + 2 πn

代入

n = 1

arcsin (- \frac{6 - 2}{4}) + 2 π 1

对于

2 cos (x) - 23 sin (x) = 8

代入

x = arcsin (- \frac{6 - 2}{4}) + 2 π 1

2 cos (arcsin (- \frac{6 - 2}{4}) + 2 π 1) - 23 sin (arcsin (- \frac{6 - 2}{4}) + 2 π 1) = 8

整理后得

2.82842 \dots = 2.82842 \dots

\Rightarrow 真

检验

π + arcsin (\frac{6 - 2}{4}) + 2 πn

的解:

假

π + arcsin (\frac{6 - 2}{4}) + 2 πn

代入

n = 1

π + arcsin (\frac{6 - 2}{4}) + 2 π 1

对于

2 cos (x) - 23 sin (x) = 8

代入

x = π + arcsin (\frac{6 - 2}{4}) + 2 π 1

2 cos (π + arcsin (\frac{6 - 2}{4}) + 2 π 1) - 23 sin (π + arcsin (\frac{6 - 2}{4}) + 2 π 1) = 8

整理后得

- 1.03527 \dots = 2.82842 \dots

\Rightarrow 假

x = π + arcsin (\frac{6 + 2}{4}) + 2 πn, x = arcsin (- \frac{6 - 2}{4}) + 2 πn

以小数形式表示解

x = π + 1.30899 \dots + 2 πn, x = - 0.26179 \dots + 2 πn

作图

流行的例子

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tan(x)-3^{1/2}=0

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