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受欢迎的三角函数 >

-cos(x)-sin(x)=1

解答

- cos (x) - sin (x) = 1

解答

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

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

- cos (x) - sin (x) = 1

两边加上

sin (x)

- cos (x) = 1 + sin (x)

两边进行平方

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

两边减去

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

cos^{2} (x) - 1 - 2 sin (x) - sin^{2} (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

化简

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

- 2 sin (x) - 2 sin^{2} (x) = 0

用替代法求解

- 2 sin (x) - 2 sin^{2} (x) = 0

令

: sin (x) = u

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

- 2 u - 2 u^{2} = 0 : u = - 1, u = 0

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = - 2, c = 0

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

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

(- 2)^{2} - 4 (- 2) \cdot 0 = 2

(- 2)^{2} - 4 (- 2) \cdot 0

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 2 \cdot 0

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 2

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

将解分隔开

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

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

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

去除括号:

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

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

数字相加:

2 + 2 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{- 4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

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

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

去除括号:

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

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

数字相减:

2 - 2 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{- 4}

使用分式法则:

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

= - \frac{0}{4}

使用法则

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

= - 0

= 0

二次方程组的解是:

u = - 1, u = 0

u = sin (x)

代回

sin (x) = - 1, sin (x) = 0

sin (x) = - 1, sin (x) = 0

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

合并所有解

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

将解代入原方程进行验证

将它们代入

- cos (x) - sin (x) = 1

检验解是否符合

去除与方程不符的解。

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

- cos (x) - sin (x) = 1

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

检验

2 πn

的解:

假

2 πn

代入

n = 1

2 π 1

对于

- cos (x) - sin (x) = 1

代入

x = 2 π 1

- cos (2 π 1) - sin (2 π 1) = 1

整理后得

- 1 = 1

\Rightarrow 假

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

- cos (x) - sin (x) = 1

代入

x = π + 2 π 1

- cos (π + 2 π 1) - sin (π + 2 π 1) = 1

整理后得

1 = 1

\Rightarrow 真

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

作图

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