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受欢迎的三角函数 >

2cot(x)cos(x)+7=7csc(x)

解答

2 cot (x) cos (x) + 7 = 7 csc (x)

解答

x = \frac{π}{2} + 2 πn

+1

度数

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

求解步骤

2 cot (x) cos (x) + 7 = 7 csc (x)

两边减去

7 csc (x)

2 cot (x) cos (x) + 7 - 7 csc (x) = 0

用 sin, cos 表示

2 \cdot \frac{cos ( x )}{sin ( x )} cos (x) + 7 - 7 \cdot \frac{1}{sin ( x )} = 0

化简

2 \cdot \frac{cos ( x )}{sin ( x )} cos (x) + 7 - 7 \cdot \frac{1}{sin ( x )} : \frac{2 cos ^{2} ( x ) - 7 + 7 sin ( x )}{sin ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )} cos (x) + 7 - 7 \cdot \frac{1}{sin ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )} cos (x) = \frac{2 cos ^{2} ( x )}{sin ( x )}

2 \cdot \frac{cos ( x )}{sin ( x )} cos (x)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{cos ( x ) \cdot 2 cos ( x )}{sin ( x )}

cos (x) \cdot 2 cos (x) = 2 cos^{2} (x)

cos (x) \cdot 2 cos (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= 2 cos^{2} (x)

= \frac{2 cos ^{2} ( x )}{sin ( x )}

7 \cdot \frac{1}{sin ( x )} = \frac{7}{sin ( x )}

7 \cdot \frac{1}{sin ( x )}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 7}{sin ( x )}

数字相乘:

1 \cdot 7 = 7

= \frac{7}{sin ( x )}

= \frac{2 cos ^{2} ( x )}{sin ( x )} + 7 - \frac{7}{sin ( x )}

合并分式

\frac{2 cos ^{2} ( x )}{sin ( x )} - \frac{7}{sin ( x )} : \frac{2 cos ^{2} ( x ) - 7}{sin ( x )}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 cos ^{2} ( x ) - 7}{sin ( x )}

= \frac{2 cos ^{2} ( x ) - 7}{sin ( x )} + 7

将项转换为分式:

7 = \frac{7 sin ( x )}{sin ( x )}

= \frac{2 cos ^{2} ( x ) - 7}{sin ( x )} + \frac{7 sin ( x )}{sin ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 cos ^{2} ( x ) - 7 + 7 sin ( x )}{sin ( x )}

\frac{2 cos ^{2} ( x ) - 7 + 7 sin ( x )}{sin ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cos^{2} (x) - 7 + 7 sin (x) = 0

两边减去

7 sin (x)

2 cos^{2} (x) - 7 = - 7 sin (x)

两边进行平方

(2 cos^{2} (x) - 7)^{2} = (- 7 sin (x))^{2}

两边减去

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

(2 cos^{2} (x) - 7)^{2} - 49 sin^{2} (x) = 0

分解

(2 cos^{2} (x) - 7)^{2} - 49 sin^{2} (x) : (2 cos^{2} (x) - 7 + 7 sin (x)) (2 cos^{2} (x) - 7 - 7 sin (x))

(2 cos^{2} (x) - 7)^{2} - 49 sin^{2} (x)

将

(2 cos^{2} (x) - 7)^{2} - 49 sin^{2} (x)

改写为

(2 cos^{2} (x) - 7)^{2} - (7 sin (x))^{2}

(2 cos^{2} (x) - 7)^{2} - 49 sin^{2} (x)

将

49

改写为

7^{2}

= (2 cos^{2} (x) - 7)^{2} - 7^{2} sin^{2} (x)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

7^{2} sin^{2} (x) = (7 sin (x))^{2}

= (2 cos^{2} (x) - 7)^{2} - (7 sin (x))^{2}

= (2 cos^{2} (x) - 7)^{2} - (7 sin (x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos^{2} (x) - 7)^{2} - (7 sin (x))^{2} = ((2 cos^{2} (x) - 7) + 7 sin (x)) ((2 cos^{2} (x) - 7) - 7 sin (x))

= ((2 cos^{2} (x) - 7) + 7 sin (x)) ((2 cos^{2} (x) - 7) - 7 sin (x))

整理后得

= (2 cos^{2} (x) + 7 sin (x) - 7) (2 cos^{2} (x) - 7 sin (x) - 7)

(2 cos^{2} (x) - 7 + 7 sin (x)) (2 cos^{2} (x) - 7 - 7 sin (x)) = 0

分别求解每个部分

2 cos^{2} (x) - 7 + 7 sin (x) = 0 or 2 cos^{2} (x) - 7 - 7 sin (x) = 0

2 cos^{2} (x) - 7 + 7 sin (x) = 0 : x = \frac{π}{2} + 2 πn

2 cos^{2} (x) - 7 + 7 sin (x) = 0

使用三角恒等式改写

- 7 + 2 cos^{2} (x) + 7 sin (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)

= - 7 + 2 (1 - sin^{2} (x)) + 7 sin (x)

化简

- 7 + 2 (1 - sin^{2} (x)) + 7 sin (x) : 7 sin (x) - 2 sin^{2} (x) - 5

- 7 + 2 (1 - sin^{2} (x)) + 7 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)

= - 7 + 2 - 2 sin^{2} (x) + 7 sin (x)

数字相加/相减:

- 7 + 2 = - 5

= 7 sin (x) - 2 sin^{2} (x) - 5

= 7 sin (x) - 2 sin^{2} (x) - 5

- 5 - 2 sin^{2} (x) + 7 sin (x) = 0

用替代法求解

- 5 - 2 sin^{2} (x) + 7 sin (x) = 0

令

: sin (x) = u

- 5 - 2 u^{2} + 7 u = 0

- 5 - 2 u^{2} + 7 u = 0 : u = 1, u = \frac{5}{2}

- 5 - 2 u^{2} + 7 u = 0

改写成标准形式

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

- 2 u^{2} + 7 u - 5 = 0

使用求根公式求解

- 2 u^{2} + 7 u - 5 = 0

二次方程求根公式:

若

a = - 2, b = 7, c = - 5

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

7^{2} - 4 (- 2) (- 5) = 3

7^{2} - 4 (- 2) (- 5)

使用法则

- (- a) = a

= 7^{2} - 4 \cdot 2 \cdot 5

数字相乘:

4 \cdot 2 \cdot 5 = 40

= 7^{2} - 40

7^{2} = 49

= 49 - 40

数字相减:

49 - 40 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 7 \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 7 + 3}{2 ( - 2 )}, u_{2} = \frac{- 7 - 3}{2 ( - 2 )}

u = \frac{- 7 + 3}{2 ( - 2 )} : 1

\frac{- 7 + 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 7 + 3}{- 2 \cdot 2}

数字相加/相减:

- 7 + 3 = - 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{- 7 - 3}{2 ( - 2 )} : \frac{5}{2}

\frac{- 7 - 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 7 - 3}{- 2 \cdot 2}

数字相减:

- 7 - 3 = - 10

= \frac{- 10}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 10}{- 4}

使用分式法则:

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

= \frac{10}{4}

约分:

2

= \frac{5}{2}

二次方程组的解是:

u = 1, u = \frac{5}{2}

u = sin (x)

代回

sin (x) = 1, sin (x) = \frac{5}{2}

sin (x) = 1, sin (x) = \frac{5}{2}

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

sin (x) = \frac{5}{2} :

无解

sin (x) = \frac{5}{2}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{π}{2} + 2 πn

2 cos^{2} (x) - 7 - 7 sin (x) = 0 : x = \frac{3 π}{2} + 2 πn

2 cos^{2} (x) - 7 - 7 sin (x) = 0

使用三角恒等式改写

- 7 + 2 cos^{2} (x) - 7 sin (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)

= - 7 + 2 (1 - sin^{2} (x)) - 7 sin (x)

化简

- 7 + 2 (1 - sin^{2} (x)) - 7 sin (x) : - 2 sin^{2} (x) - 7 sin (x) - 5

- 7 + 2 (1 - sin^{2} (x)) - 7 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)

= - 7 + 2 - 2 sin^{2} (x) - 7 sin (x)

数字相加/相减:

- 7 + 2 = - 5

= - 2 sin^{2} (x) - 7 sin (x) - 5

= - 2 sin^{2} (x) - 7 sin (x) - 5

- 5 - 2 sin^{2} (x) - 7 sin (x) = 0

用替代法求解

- 5 - 2 sin^{2} (x) - 7 sin (x) = 0

令

: sin (x) = u

- 5 - 2 u^{2} - 7 u = 0

- 5 - 2 u^{2} - 7 u = 0 : u = - \frac{5}{2}, u = - 1

- 5 - 2 u^{2} - 7 u = 0

改写成标准形式

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

- 2 u^{2} - 7 u - 5 = 0

使用求根公式求解

- 2 u^{2} - 7 u - 5 = 0

二次方程求根公式:

若

a = - 2, b = - 7, c = - 5

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 2 ) ( - 5 )}{2 ( - 2 )}

(- 7)^{2} - 4 (- 2) (- 5) = 3

(- 7)^{2} - 4 (- 2) (- 5)

使用法则

- (- a) = a

= (- 7)^{2} - 4 \cdot 2 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 2 \cdot 5

数字相乘:

4 \cdot 2 \cdot 5 = 40

= 7^{2} - 40

7^{2} = 49

= 49 - 40

数字相减:

49 - 40 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 7 ) \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 7 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 7 ) + 3}{2 ( - 2 )} : - \frac{5}{2}

\frac{- ( - 7 ) + 3}{2 ( - 2 )}

去除括号:

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

= \frac{7 + 3}{- 2 \cdot 2}

数字相加:

7 + 3 = 10

= \frac{10}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{10}{- 4}

使用分式法则:

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

= - \frac{10}{4}

约分:

2

= - \frac{5}{2}

u = \frac{- ( - 7 ) - 3}{2 ( - 2 )} : - 1

\frac{- ( - 7 ) - 3}{2 ( - 2 )}

去除括号:

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

= \frac{7 - 3}{- 2 \cdot 2}

数字相减:

7 - 3 = 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

二次方程组的解是:

u = - \frac{5}{2}, u = - 1

u = sin (x)

代回

sin (x) = - \frac{5}{2}, sin (x) = - 1

sin (x) = - \frac{5}{2}, sin (x) = - 1

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

无解

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

- 1 \leq sin (x) \leq 1

无解

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 = \frac{3 π}{2} + 2 πn

合并所有解

x = \frac{π}{2} + 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

2 cot (x) cos (x) + 7 = 7 csc (x)

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

2 cot (x) cos (x) + 7 = 7 csc (x)

代入

x = \frac{π}{2} + 2 π 1

2 cot (\frac{π}{2} + 2 π 1) cos (\frac{π}{2} + 2 π 1) + 7 = 7 csc (\frac{π}{2} + 2 π 1)

整理后得

7 = 7

\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

2 cot (x) cos (x) + 7 = 7 csc (x)

代入

x = \frac{3 π}{2} + 2 π 1

2 cot (\frac{3 π}{2} + 2 π 1) cos (\frac{3 π}{2} + 2 π 1) + 7 = 7 csc (\frac{3 π}{2} + 2 π 1)

整理后得

7 = - 7

\Rightarrow 假

x = \frac{π}{2} + 2 πn

作图

流行的例子

2cos(2x-pi/3)=1

2 cos (2 x - \frac{π}{3}) = 1

cos(2θ)= 1/(sqrt(2))

cos (2 θ) = \frac{1}{2}

1 + 4 sin^{2} (θ) = 2

arcsin(x)-arctan(sqrt(3))=-pi/6

arcsin (x) - arctan (3) = - \frac{π}{6}

tan(x)=2sqrt(2)

tan (x) = 22