zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

x = \frac{7 π}{4} + 2 πn

+1

度数

x = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

2 sin (x)

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

两边进行平方

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

两边减去

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

2 cos^{2} (x) - 4 - 42 sin (x) - 2 sin^{2} (x) = 0

使用三角恒等式改写

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

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

化简

- 4 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 4 sin (x) 2 : - 4 sin^{2} (x) - 42 sin (x) - 2

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

= - 4 + 2 (1 - sin^{2} (x)) - 2 sin^{2} (x) - 42 sin (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

2 \cdot 1 = 2

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

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

化简

- 4 + 2 - 2 sin^{2} (x) - 2 sin^{2} (x) - 4 sin (x) 2 : - 4 sin^{2} (x) - 42 sin (x) - 2

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

同类项相加:

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

= - 4 + 2 - 4 sin^{2} (x) - 42 sin (x)

数字相加/相减:

- 4 + 2 = - 2

= - 4 sin^{2} (x) - 42 sin (x) - 2

= - 4 sin^{2} (x) - 42 sin (x) - 2

= - 4 sin^{2} (x) - 42 sin (x) - 2

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

用替代法求解

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

令

: sin (x) = u

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

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

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

改写成标准形式

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

- 4 u^{2} - 42 u - 2 = 0

使用求根公式求解

- 4 u^{2} - 42 u - 2 = 0

二次方程求根公式:

若

a = - 4, b = - 42, c = - 2

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

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

(- 42)^{2} - 4 (- 4) (- 2) = 0

(- 42)^{2} - 4 (- 4) (- 2)

使用法则

- (- a) = a

= (- 42)^{2} - 4 \cdot 4 \cdot 2

(- 42)^{2} = 4^{2} \cdot 2

(- 42)^{2}

使用指数法则:

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

若

n

是偶数

(- 42)^{2} = (42)^{2}

= (42)^{2}

使用指数法则:

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

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

(2)^{2} : 2

使用根式运算法则:

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

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

使用指数法则:

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

= 2^{\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

= 2

= 4^{2} \cdot 2

4 \cdot 4 \cdot 2 = 32

4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 32

= 4^{2} \cdot 2 - 32

4^{2} \cdot 2 = 32

4^{2} \cdot 2

4^{2} = 16

= 16 \cdot 2

数字相乘:

16 \cdot 2 = 32

= 32

= 32 - 32

数字相减:

32 - 32 = 0

= 0

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

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

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

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

去除括号:

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

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

数字相乘:

2 \cdot 4 = 8

= \frac{4 2}{- 8}

使用分式法则:

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

= - \frac{4 2}{8}

约分:

4

= - \frac{2}{2}

u = - \frac{2}{2}

二次方程组的解是:

u = - \frac{2}{2}

u = sin (x)

代回

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

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

sin (x) = - \frac{2}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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

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

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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

合并所有解

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

\frac{5 π}{4} + 2 πn

的解:

假

\frac{5 π}{4} + 2 πn

代入

n = 1

\frac{5 π}{4} + 2 π 1

对于

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

代入

x = \frac{5 π}{4} + 2 π 1

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

整理后得

0 = 2

\Rightarrow 假

检验

\frac{7 π}{4} + 2 πn

的解:

真

\frac{7 π}{4} + 2 πn

代入

n = 1

\frac{7 π}{4} + 2 π 1

对于

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

代入

x = \frac{7 π}{4} + 2 π 1

2 cos (\frac{7 π}{4} + 2 π 1) - 2 sin (\frac{7 π}{4} + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

x = \frac{7 π}{4} + 2 πn

作图

流行的例子

sin^2(x)-3sin(x)=-2

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

solvefor x,f=arctan(x/(sqrt(1-x^2)))

so l v e f or x, f = arctan (\frac{x}{1 - x ^{2}})

sin(2x)=((8m-2))/5

sin (2 x) = \frac{( 8 m - 2 )}{5}

csc(3x)=sin(3x)

csc (3 x) = sin (3 x)

sin^2(x)=((10m-7))/9

sin^{2} (x) = \frac{( 10 m - 7 )}{9}