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受欢迎的三角函数 >

3tan^2(x)= 8/(sin^2(x))

解答

3 tan^{2} (x) = \frac{8}{sin ^{2} ( x )}

解答

x = 1.07640 \dots + 2 πn, x = 2 π - 1.07640 \dots + 2 πn, x = 2.06518 \dots + 2 πn, x = - 2.06518 \dots + 2 πn

+1

度数

x = 61.67333 \dots^{\circ} + 36 0^{\circ} n, x = 298.32666 \dots^{\circ} + 36 0^{\circ} n, x = 118.32666 \dots^{\circ} + 36 0^{\circ} n, x = - 118.32666 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 tan^{2} (x) = \frac{8}{sin ^{2} ( x )}

两边减去

\frac{8}{sin ^{2} ( x )}

3 tan^{2} (x) - \frac{8}{sin ^{2} ( x )} = 0

化简

3 tan^{2} (x) - \frac{8}{sin ^{2} ( x )} : \frac{3 tan ^{2} ( x ) sin ^{2} ( x ) - 8}{sin ^{2} ( x )}

3 tan^{2} (x) - \frac{8}{sin ^{2} ( x )}

将项转换为分式:

3 tan^{2} (x) = \frac{3 tan ^{2} ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

= \frac{3 tan ^{2} ( x ) sin ^{2} ( x )}{sin ^{2} ( x )} - \frac{8}{sin ^{2} ( 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{3 tan ^{2} ( x ) sin ^{2} ( x ) - 8}{sin ^{2} ( x )}

\frac{3 tan ^{2} ( x ) sin ^{2} ( x ) - 8}{sin ^{2} ( x )} = 0

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

3 tan^{2} (x) sin^{2} (x) - 8 = 0

使用三角恒等式改写

- 8 + 3 sin^{2} (x) tan^{2} (x)

tan (x) = - tan (π - x)

tan (x)

利用以下特性:

tan (θ) = - tan (- θ)

tan (x) = - tan (- x)

= - tan (- x)

使用周期

tan

:

tan (π + θ) = tan (θ)

- tan (- x) = - tan (π - x)

= - tan (π - x)

= - 8 + 3 sin^{2} (x) (- tan (π - x))^{2}

使用指数法则:

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

若

n

是偶数

(- tan (- x + π))^{2} = tan^{2} (π - x)

= - 8 + 3 sin^{2} (x) tan^{2} (π - x)

- 8 + 3 sin^{2} (x) tan^{2} (- x + π) = 0

分解

- 8 + 3 sin^{2} (x) tan^{2} (- x + π) : (3 sin (x) tan (- x + π) + 22) (3 sin (x) tan (- x + π) - 22)

- 8 + 3 sin^{2} (x) tan^{2} (- x + π)

将

- 8 + 3 sin^{2} (x) tan^{2} (- x + π)

改写为

- (8)^{2} + (3 sin (x) tan (- x + π))^{2}

- 8 + 3 sin^{2} (x) tan^{2} (- x + π)

使用根式运算法则:

a = (a)^{2}

8 = (8)^{2}

= - (8)^{2} + 3 sin^{2} (x) tan^{2} (- x + π)

使用根式运算法则:

a = (a)^{2}

3 = (3)^{2}

= - (8)^{2} + (3)^{2} sin^{2} (x) tan^{2} (- x + π)

使用指数法则:

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

(3)^{2} sin^{2} (x) tan^{2} (- x + π) = (3 sin (x) tan (- x + π))^{2}

= - (8)^{2} + (3 sin (x) tan (- x + π))^{2}

= - (8)^{2} + (3 sin (x) tan (- x + π))^{2}

使用平方差公式:

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

- (8)^{2} + (3 sin (x) tan (- x + π))^{2} = (3 sin (x) tan (- x + π) + 8) (3 sin (x) tan (- x + π) - 8)

= (3 sin (x) tan (- x + π) + 8) (3 sin (x) tan (- x + π) - 8)

整理后得

= (3 sin (x) tan (- x + π) + 22) (3 sin (x) tan (- x + π) - 22)

(3 sin (x) tan (- x + π) + 22) (3 sin (x) tan (- x + π) - 22) = 0

分别求解每个部分

3 sin (x) tan (- x + π) + 22 = 0 or 3 sin (x) tan (- x + π) - 22 = 0

3 sin (x) tan (- x + π) + 22 = 0 : x = arccos (\frac{3 ( - 2 + 5 )}{3}) + 2 πn, x = 2 π - arccos (\frac{3 ( - 2 + 5 )}{3}) + 2 πn

3 sin (x) tan (- x + π) + 22 = 0

使用三角恒等式改写

3 sin (x) tan (- x + π) + 22 = 0

使用三角恒等式改写

tan (- x + π)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( - x + π )}{cos ( - x + π )}

使用角差恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π - x )}

使用角差恒等式:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

化简

\frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )} : - \frac{sin ( x )}{cos ( x )}

\frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

sin (π) cos (x) - cos (π) sin (x) = sin (x)

sin (π) cos (x) - cos (π) sin (x)

sin (π) cos (x) = 0

sin (π) cos (x)

化简

sin (π) : 0

sin (π)

使用以下普通恒等式:

sin (π) = 0

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}

= 0

= 0 \cdot cos (x)

使用法则

0 \cdot a = 0

= 0

cos (π) sin (x) = - sin (x)

cos (π) sin (x)

化简

cos (π) : - 1

cos (π)

使用以下普通恒等式:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= - 1 \cdot sin (x)

乘以:

1 \cdot sin (x) = sin (x)

= - sin (x)

= 0 - (- sin (x))

整理后得

= sin (x)

= \frac{sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

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

cos (π) cos (x) + sin (π) sin (x)

cos (π) cos (x) = - cos (x)

cos (π) cos (x)

化简

cos (π) : - 1

cos (π)

使用以下普通恒等式:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= - 1 \cdot cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (π) sin (x) = 0

sin (π) sin (x)

化简

sin (π) : 0

sin (π)

使用以下普通恒等式:

sin (π) = 0

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}

= 0

= 0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

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

使用分式法则:

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

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

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

3 sin (x) (- \frac{sin ( x )}{cos ( x )}) + 22 = 0

化简

3 sin (x) (- \frac{sin ( x )}{cos ( x )}) + 22 : - \frac{3 sin ^{2} ( x )}{cos ( x )} + 22

3 sin (x) (- \frac{sin ( x )}{cos ( x )}) + 22

去除括号:

(- a) = - a

= - 3 sin (x) \frac{sin ( x )}{cos ( x )} + 22

3 sin (x) \frac{sin ( x )}{cos ( x )} = \frac{3 sin ^{2} ( x )}{cos ( x )}

3 sin (x) \frac{sin ( x )}{cos ( x )}

分式相乘:

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

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

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

sin (x) 3 sin (x)

使用指数法则:

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

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

= 3 sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 3 sin^{2} (x)

= \frac{3 sin ^{2} ( x )}{cos ( x )}

= - \frac{3 sin ^{2} ( x )}{cos ( x )} + 22

- \frac{3 sin ^{2} ( x )}{cos ( x )} + 22 = 0

- \frac{3 sin ^{2} ( x )}{cos ( x )} + 22 = 0

化简

- \frac{3 sin ^{2} ( x )}{cos ( x )} + 22 : \frac{- 3 sin ^{2} ( x ) + 2 2 cos ( x )}{cos ( x )}

- \frac{3 sin ^{2} ( x )}{cos ( x )} + 22

将项转换为分式:

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

= - \frac{3 sin ^{2} ( x )}{cos ( x )} + \frac{2 2 cos ( x )}{cos ( 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{- 3 sin ^{2} ( x ) + 2 2 cos ( x )}{cos ( x )}

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

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

- 3 sin^{2} (x) + 22 cos (x) = 0

使用三角恒等式改写

- sin^{2} (x) 3 + 2 cos (x) 2

使用毕达哥拉斯恒等式:

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

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

= - (1 - cos^{2} (x)) 3 + 2 cos (x) 2

- (1 - cos^{2} (x)) 3 + 2 cos (x) 2 = 0

用替代法求解

- (1 - cos^{2} (x)) 3 + 2 cos (x) 2 = 0

令

: cos (x) = u

- (1 - u^{2}) 3 + 2 u 2 = 0

- (1 - u^{2}) 3 + 2 u 2 = 0 : u = \frac{3 ( - 2 + 5 )}{3}, u = - \frac{3 ( 2 + 5 )}{3}

- (1 - u^{2}) 3 + 2 u 2 = 0

展开

- (1 - u^{2}) 3 + 2 u 2 : - 3 + 3 u^{2} + 22 u

- (1 - u^{2}) 3 + 2 u 2

= - 3 (1 - u^{2}) + 22 u

乘开

- 3 (1 - u^{2}) : - 3 + 3 u^{2}

- 3 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = u^{2}

= - 3 \cdot 1 - (- 3) u^{2}

使用加减运算法则

- (- a) = a

= - 1 \cdot 3 + 3 u^{2}

乘以:

1 \cdot 3 = 3

= - 3 + 3 u^{2}

= - 3 + 3 u^{2} + 2 u 2

= - 3 + 3 u^{2} + 22 u

- 3 + 3 u^{2} + 22 u = 0

改写成标准形式

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

3 u^{2} + 22 u - 3 = 0

使用求根公式求解

3 u^{2} + 22 u - 3 = 0

二次方程求根公式:

若

a = 3, b = 22, c = - 3

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

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

(22)^{2} - 43 (- 3) = 25

(22)^{2} - 43 (- 3)

使用法则

- (- a) = a

= (22)^{2} + 433

(22)^{2} = 2^{3}

(22)^{2}

使用指数法则:

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

= 2^{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

= 2^{2} \cdot 2

使用指数法则:

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

2^{2} \cdot 2 = 2^{2 + 1}

= 2^{2 + 1}

数字相加:

2 + 1 = 3

= 2^{3}

433 = 12

433

使用根式运算法则:

a a = a

33 = 3

= 4 \cdot 3

数字相乘:

4 \cdot 3 = 12

= 12

= 2^{3} + 12

2^{3} = 8

= 8 + 12

数字相加:

8 + 12 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

u_{1, 2} = \frac{- 2 2 \pm 2 5}{2 3}

将解分隔开

u_{1} = \frac{- 2 2 + 2 5}{2 3}, u_{2} = \frac{- 2 2 - 2 5}{2 3}

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

\frac{- 2 2 + 2 5}{2 3}

因式分解出通项

2

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

数字相除:

\frac{2}{2} = 1

= \frac{- 2 + 5}{3}

\frac{- 2 + 5}{3}

有理化:

\frac{3 ( 5 - 2 )}{3}

\frac{- 2 + 5}{3}

乘以共轭根式

\frac{3}{3}

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

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

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

= \frac{3 ( 5 - 2 )}{3}

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

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

\frac{- 2 2 - 2 5}{2 3}

因式分解出通项

2

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

数字相除:

\frac{2}{2} = 1

= - \frac{2 + 5}{3}

- \frac{2 + 5}{3}

有理化:

- \frac{3 ( 2 + 5 )}{3}

- \frac{2 + 5}{3}

乘以共轭根式

\frac{3}{3}

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

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

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

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

二次方程组的解是:

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

u = cos (x)

代回

cos (x) = \frac{3 ( - 2 + 5 )}{3}, cos (x) = - \frac{3 ( 2 + 5 )}{3}

cos (x) = \frac{3 ( - 2 + 5 )}{3}, cos (x) = - \frac{3 ( 2 + 5 )}{3}

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

cos (x) = \frac{3 ( - 2 + 5 )}{3}

使用反三角函数性质

cos (x) = \frac{3 ( - 2 + 5 )}{3}

cos (x) = \frac{3 ( - 2 + 5 )}{3}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

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

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

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

无解

cos (x) = - \frac{3 ( 2 + 5 )}{3}

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

3 sin (x) tan (- x + π) - 22 = 0 : x = arccos (\frac{3 ( 2 - 5 )}{3}) + 2 πn, x = - arccos (\frac{3 ( 2 - 5 )}{3}) + 2 πn

3 sin (x) tan (- x + π) - 22 = 0

使用三角恒等式改写

3 sin (x) tan (- x + π) - 22 = 0

使用三角恒等式改写

tan (- x + π)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= \frac{sin ( - x + π )}{cos ( - x + π )}

使用角差恒等式:

sin (s - t) = sin (s) cos (t) - cos (s) sin (t)

= \frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π - x )}

使用角差恒等式:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= \frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

化简

\frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )} : - \frac{sin ( x )}{cos ( x )}

\frac{sin ( π ) cos ( x ) - cos ( π ) sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

sin (π) cos (x) - cos (π) sin (x) = sin (x)

sin (π) cos (x) - cos (π) sin (x)

sin (π) cos (x) = 0

sin (π) cos (x)

化简

sin (π) : 0

sin (π)

使用以下普通恒等式:

sin (π) = 0

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}

= 0

= 0 \cdot cos (x)

使用法则

0 \cdot a = 0

= 0

cos (π) sin (x) = - sin (x)

cos (π) sin (x)

化简

cos (π) : - 1

cos (π)

使用以下普通恒等式:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= - 1 \cdot sin (x)

乘以:

1 \cdot sin (x) = sin (x)

= - sin (x)

= 0 - (- sin (x))

整理后得

= sin (x)

= \frac{sin ( x )}{cos ( π ) cos ( x ) + sin ( π ) sin ( x )}

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

cos (π) cos (x) + sin (π) sin (x)

cos (π) cos (x) = - cos (x)

cos (π) cos (x)

化简

cos (π) : - 1

cos (π)

使用以下普通恒等式:

cos (π) = (- 1)

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - 1

= - 1 \cdot cos (x)

乘以:

1 \cdot cos (x) = cos (x)

= - cos (x)

= - cos (x) + sin (π) sin (x)

sin (π) sin (x) = 0

sin (π) sin (x)

化简

sin (π) : 0

sin (π)

使用以下普通恒等式:

sin (π) = 0

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}

= 0

= 0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

= - cos (x) + 0

- cos (x) + 0 = - cos (x)

= - cos (x)

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

使用分式法则:

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

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

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

3 sin (x) (- \frac{sin ( x )}{cos ( x )}) - 22 = 0

化简

3 sin (x) (- \frac{sin ( x )}{cos ( x )}) - 22 : - \frac{3 sin ^{2} ( x )}{cos ( x )} - 22

3 sin (x) (- \frac{sin ( x )}{cos ( x )}) - 22

去除括号:

(- a) = - a

= - 3 sin (x) \frac{sin ( x )}{cos ( x )} - 22

3 sin (x) \frac{sin ( x )}{cos ( x )} = \frac{3 sin ^{2} ( x )}{cos ( x )}

3 sin (x) \frac{sin ( x )}{cos ( x )}

分式相乘:

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

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

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

sin (x) 3 sin (x)

使用指数法则:

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

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

= 3 sin^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 3 sin^{2} (x)

= \frac{3 sin ^{2} ( x )}{cos ( x )}

= - \frac{3 sin ^{2} ( x )}{cos ( x )} - 22

- \frac{3 sin ^{2} ( x )}{cos ( x )} - 22 = 0

- \frac{3 sin ^{2} ( x )}{cos ( x )} - 22 = 0

化简

- \frac{3 sin ^{2} ( x )}{cos ( x )} - 22 : \frac{- 3 sin ^{2} ( x ) - 2 2 cos ( x )}{cos ( x )}

- \frac{3 sin ^{2} ( x )}{cos ( x )} - 22

将项转换为分式:

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

= - \frac{3 sin ^{2} ( x )}{cos ( x )} - \frac{2 2 cos ( x )}{cos ( 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{- 3 sin ^{2} ( x ) - 2 2 cos ( x )}{cos ( x )}

\frac{- 3 sin ^{2} ( x ) - 2 2 cos ( x )}{cos ( x )} = 0

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

- 3 sin^{2} (x) - 22 cos (x) = 0

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= - (1 - cos^{2} (x)) 3 - 2 cos (x) 2

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

用替代法求解

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

令

: cos (x) = u

- (1 - u^{2}) 3 - 2 u 2 = 0

- (1 - u^{2}) 3 - 2 u 2 = 0 : u = \frac{3 ( 2 + 5 )}{3}, u = \frac{3 ( 2 - 5 )}{3}

- (1 - u^{2}) 3 - 2 u 2 = 0

展开

- (1 - u^{2}) 3 - 2 u 2 : - 3 + 3 u^{2} - 22 u

- (1 - u^{2}) 3 - 2 u 2

= - 3 (1 - u^{2}) - 22 u

乘开

- 3 (1 - u^{2}) : - 3 + 3 u^{2}

- 3 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = u^{2}

= - 3 \cdot 1 - (- 3) u^{2}

使用加减运算法则

- (- a) = a

= - 1 \cdot 3 + 3 u^{2}

乘以:

1 \cdot 3 = 3

= - 3 + 3 u^{2}

= - 3 + 3 u^{2} - 2 u 2

= - 3 + 3 u^{2} - 22 u

- 3 + 3 u^{2} - 22 u = 0

改写成标准形式

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

3 u^{2} - 22 u - 3 = 0

使用求根公式求解

3 u^{2} - 22 u - 3 = 0

二次方程求根公式:

若

a = 3, b = - 22, c = - 3

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

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

(- 22)^{2} - 43 (- 3) = 25

(- 22)^{2} - 43 (- 3)

使用法则

- (- a) = a

= (- 22)^{2} + 433

(- 22)^{2} = 2^{3}

(- 22)^{2}

使用指数法则:

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

若

n

是偶数

(- 22)^{2} = (22)^{2}

= (22)^{2}

使用指数法则:

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

= 2^{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

= 2^{2} \cdot 2

使用指数法则:

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

2^{2} \cdot 2 = 2^{2 + 1}

= 2^{2 + 1}

数字相加:

2 + 1 = 3

= 2^{3}

433 = 12

433

使用根式运算法则:

a a = a

33 = 3

= 4 \cdot 3

数字相乘:

4 \cdot 3 = 12

= 12

= 2^{3} + 12

2^{3} = 8

= 8 + 12

数字相加:

8 + 12 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

u_{1, 2} = \frac{- ( - 2 2 ) \pm 2 5}{2 3}

将解分隔开

u_{1} = \frac{- ( - 2 2 ) + 2 5}{2 3}, u_{2} = \frac{- ( - 2 2 ) - 2 5}{2 3}

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

\frac{- ( - 2 2 ) + 2 5}{2 3}

使用法则

- (- a) = a

= \frac{2 2 + 2 5}{2 3}

因式分解出通项

2

= \frac{2 ( 2 + 5 )}{2 3}

数字相除:

\frac{2}{2} = 1

= \frac{2 + 5}{3}

\frac{2 + 5}{3}

有理化:

\frac{3 ( 2 + 5 )}{3}

\frac{2 + 5}{3}

乘以共轭根式

\frac{3}{3}

= \frac{( 2 + 5 ) 3}{3 3}

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{3 ( 2 + 5 )}{3}

= \frac{3 ( 2 + 5 )}{3}

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

\frac{- ( - 2 2 ) - 2 5}{2 3}

使用法则

- (- a) = a

= \frac{2 2 - 2 5}{2 3}

因式分解出通项

2

= \frac{2 ( 2 - 5 )}{2 3}

数字相除:

\frac{2}{2} = 1

= \frac{2 - 5}{3}

\frac{2 - 5}{3}

有理化:

\frac{3 ( 2 - 5 )}{3}

\frac{2 - 5}{3}

乘以共轭根式

\frac{3}{3}

= \frac{( 2 - 5 ) 3}{3 3}

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= \frac{3 ( 2 - 5 )}{3}

= \frac{3 ( 2 - 5 )}{3}

二次方程组的解是:

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

u = cos (x)

代回

cos (x) = \frac{3 ( 2 + 5 )}{3}, cos (x) = \frac{3 ( 2 - 5 )}{3}

cos (x) = \frac{3 ( 2 + 5 )}{3}, cos (x) = \frac{3 ( 2 - 5 )}{3}

cos (x) = \frac{3 ( 2 + 5 )}{3} :

无解

cos (x) = \frac{3 ( 2 + 5 )}{3}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{3 ( 2 - 5 )}{3} : x = arccos (\frac{3 ( 2 - 5 )}{3}) + 2 πn, x = - arccos (\frac{3 ( 2 - 5 )}{3}) + 2 πn

cos (x) = \frac{3 ( 2 - 5 )}{3}

使用反三角函数性质

cos (x) = \frac{3 ( 2 - 5 )}{3}

cos (x) = \frac{3 ( 2 - 5 )}{3}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

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

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

合并所有解

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

合并所有解

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

以小数形式表示解

x = 1.07640 \dots + 2 πn, x = 2 π - 1.07640 \dots + 2 πn, x = 2.06518 \dots + 2 πn, x = - 2.06518 \dots + 2 πn

作图

流行的例子

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