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受欢迎的三角函数 >

4+4sin(x)=8cos(x)

解答

4 + 4 sin (x) = 8 cos (x)

解答

x = 0.64350 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度数

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

4 + 4 sin (x) = 8 cos (x)

两边进行平方

(4 + 4 sin (x))^{2} = (8 cos (x))^{2}

两边减去

(8 cos (x))^{2}

(4 + 4 sin (x))^{2} - 64 cos^{2} (x) = 0

使用三角恒等式改写

(4 + 4 sin (x))^{2} - 64 cos^{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)

= (4 + 4 sin (x))^{2} - 64 (1 - sin^{2} (x))

化简

(4 + 4 sin (x))^{2} - 64 (1 - sin^{2} (x)) : 80 sin^{2} (x) + 32 sin (x) - 48

(4 + 4 sin (x))^{2} - 64 (1 - sin^{2} (x))

(4 + 4 sin (x))^{2} : 16 + 32 sin (x) + 16 sin^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 4, b = 4 sin (x)

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

化简

4^{2} + 2 \cdot 4 \cdot 4 sin (x) + (4 sin (x))^{2} : 16 + 32 sin (x) + 16 sin^{2} (x)

4^{2} + 2 \cdot 4 \cdot 4 sin (x) + (4 sin (x))^{2}

4^{2} = 16

4^{2}

4^{2} = 16

= 16

2 \cdot 4 \cdot 4 sin (x) = 32 sin (x)

2 \cdot 4 \cdot 4 sin (x)

数字相乘:

2 \cdot 4 \cdot 4 = 32

= 32 sin (x)

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

(4 sin (x))^{2}

使用指数法则:

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

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

4^{2} = 16

= 16 sin^{2} (x)

= 16 + 32 sin (x) + 16 sin^{2} (x)

= 16 + 32 sin (x) + 16 sin^{2} (x)

= 16 + 32 sin (x) + 16 sin^{2} (x) - 64 (1 - sin^{2} (x))

乘开

- 64 (1 - sin^{2} (x)) : - 64 + 64 sin^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

数字相乘:

64 \cdot 1 = 64

= - 64 + 64 sin^{2} (x)

= 16 + 32 sin (x) + 16 sin^{2} (x) - 64 + 64 sin^{2} (x)

化简

16 + 32 sin (x) + 16 sin^{2} (x) - 64 + 64 sin^{2} (x) : 80 sin^{2} (x) + 32 sin (x) - 48

16 + 32 sin (x) + 16 sin^{2} (x) - 64 + 64 sin^{2} (x)

对同类项分组

= 32 sin (x) + 16 sin^{2} (x) + 64 sin^{2} (x) + 16 - 64

同类项相加:

16 sin^{2} (x) + 64 sin^{2} (x) = 80 sin^{2} (x)

= 32 sin (x) + 80 sin^{2} (x) + 16 - 64

数字相加/相减:

16 - 64 = - 48

= 80 sin^{2} (x) + 32 sin (x) - 48

= 80 sin^{2} (x) + 32 sin (x) - 48

= 80 sin^{2} (x) + 32 sin (x) - 48

- 48 + 32 sin (x) + 80 sin^{2} (x) = 0

用替代法求解

- 48 + 32 sin (x) + 80 sin^{2} (x) = 0

令

: sin (x) = u

- 48 + 32 u + 80 u^{2} = 0

- 48 + 32 u + 80 u^{2} = 0 : u = \frac{3}{5}, u = - 1

- 48 + 32 u + 80 u^{2} = 0

改写成标准形式

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

80 u^{2} + 32 u - 48 = 0

使用求根公式求解

80 u^{2} + 32 u - 48 = 0

二次方程求根公式:

若

a = 80, b = 32, c = - 48

u_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 80 ( - 48 )}{2 \cdot 80}

u_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 \cdot 80 ( - 48 )}{2 \cdot 80}

3 2^{2} - 4 \cdot 80 (- 48) = 128

3 2^{2} - 4 \cdot 80 (- 48)

使用法则

- (- a) = a

= 3 2^{2} + 4 \cdot 80 \cdot 48

数字相乘:

4 \cdot 80 \cdot 48 = 15360

= 3 2^{2} + 15360

3 2^{2} = 1024

= 1024 + 15360

数字相加:

1024 + 15360 = 16384

= 16384

因式分解数字:

16384 = 12 8^{2}

= 12 8^{2}

使用根式运算法则:

n a^{n} = a

12 8^{2} = 128

= 128

u_{1, 2} = \frac{- 32 \pm 128}{2 \cdot 80}

将解分隔开

u_{1} = \frac{- 32 + 128}{2 \cdot 80}, u_{2} = \frac{- 32 - 128}{2 \cdot 80}

u = \frac{- 32 + 128}{2 \cdot 80} : \frac{3}{5}

\frac{- 32 + 128}{2 \cdot 80}

数字相加/相减:

- 32 + 128 = 96

= \frac{96}{2 \cdot 80}

数字相乘:

2 \cdot 80 = 160

= \frac{96}{160}

约分:

32

= \frac{3}{5}

u = \frac{- 32 - 128}{2 \cdot 80} : - 1

\frac{- 32 - 128}{2 \cdot 80}

数字相减:

- 32 - 128 = - 160

= \frac{- 160}{2 \cdot 80}

数字相乘:

2 \cdot 80 = 160

= \frac{- 160}{160}

使用分式法则:

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

= - \frac{160}{160}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = \frac{3}{5}, u = - 1

u = sin (x)

代回

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5}, sin (x) = - 1

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = \frac{3}{5}

使用反三角函数性质

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

的通解

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

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

sin (x) = - 1

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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

合并所有解

x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

将解代入原方程进行验证

将它们代入

4 + 4 sin (x) = 8 cos (x)

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{3}{5}) + 2 πn

的解:

真

arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

arcsin (\frac{3}{5}) + 2 π 1

对于

4 + 4 sin (x) = 8 cos (x)

代入

x = arcsin (\frac{3}{5}) + 2 π 1

4 + 4 sin (arcsin (\frac{3}{5}) + 2 π 1) = 8 cos (arcsin (\frac{3}{5}) + 2 π 1)

整理后得

6.4 = 6.4

\Rightarrow 真

检验

π - arcsin (\frac{3}{5}) + 2 πn

的解:

假

π - arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

π - arcsin (\frac{3}{5}) + 2 π 1

对于

4 + 4 sin (x) = 8 cos (x)

代入

x = π - arcsin (\frac{3}{5}) + 2 π 1

4 + 4 sin (π - arcsin (\frac{3}{5}) + 2 π 1) = 8 cos (π - arcsin (\frac{3}{5}) + 2 π 1)

整理后得

6.4 = - 6.4

\Rightarrow 假

检验

\frac{3 π}{2} + 2 πn

的解:

真

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

4 + 4 sin (x) = 8 cos (x)

代入

x = \frac{3 π}{2} + 2 π 1

4 + 4 sin (\frac{3 π}{2} + 2 π 1) = 8 cos (\frac{3 π}{2} + 2 π 1)

整理后得

0 = 0

\Rightarrow 真

x = arcsin (\frac{3}{5}) + 2 πn, x = \frac{3 π}{2} + 2 πn

以小数形式表示解

x = 0.64350 \dots + 2 πn, x = \frac{3 π}{2} + 2 πn

作图

流行的例子

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