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受欢迎的三角函数 >

sin(3x)+cos(2x)=0

解答

sin (3 x) + cos (2 x) = 0

解答

x = \frac{3 π}{2} + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 5 4^{\circ} + 36 0^{\circ} n, x = 12 6^{\circ} + 36 0^{\circ} n, x = - 1 8^{\circ} + 36 0^{\circ} n, x = 19 8^{\circ} + 36 0^{\circ} n

求解步骤

sin (3 x) + cos (2 x) = 0

使用三角恒等式改写

cos (2 x) + sin (3 x)

使用倍角公式:

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

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

sin (3 x) = 3 sin (x) - 4 sin^{3} (x)

sin (3 x)

使用三角恒等式改写

sin (3 x)

改写为

= sin (2 x + x)

使用角和恒等式:

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

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

使用倍角公式:

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

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

化简

cos (2 x) sin (x) + cos (x) \cdot 2 sin (x) cos (x) : sin (x) cos (2 x) + 2 cos^{2} (x) sin (x)

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

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

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

数字相加:

1 + 1 = 2

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

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

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

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

使用倍角公式:

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

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

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

乘开

(1 - 2 sin^{2} (x)) sin (x) + 2 (1 - sin^{2} (x)) sin (x) : - 4 sin^{3} (x) + 3 sin (x)

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

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

乘开

sin (x) (1 - 2 sin^{2} (x)) : sin (x) - 2 sin^{3} (x)

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : sin (x) - 2 sin^{3} (x)

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

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

1 sin (x)

乘以:

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

= sin (x)

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

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

乘开

2 sin (x) (1 - sin^{2} (x)) : 2 sin (x) - 2 sin^{3} (x)

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

使用分配律:

a (b - c) = ab - a c

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

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

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

化简

2 \cdot 1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : 2 sin (x) - 2 sin^{3} (x)

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

2 \cdot 1 \cdot sin (x) = 2 sin (x)

2 \cdot 1 sin (x)

数字相乘:

2 \cdot 1 = 2

= 2 sin (x)

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

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

使用指数法则:

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

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

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

数字相加:

2 + 1 = 3

= 2 sin^{3} (x)

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

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

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

化简

sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x) : - 4 sin^{3} (x) + 3 sin (x)

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

对同类项分组

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

同类项相加:

- 2 sin^{3} (x) - 2 sin^{3} (x) = - 4 sin^{3} (x)

= - 4 sin^{3} (x) + sin (x) + 2 sin (x)

同类项相加:

sin (x) + 2 sin (x) = 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= 1 + 3 sin (x) - 4 sin^{3} (x) - 2 sin^{2} (x)

1 - 2 sin^{2} (x) + 3 sin (x) - 4 sin^{3} (x) = 0

用替代法求解

1 - 2 sin^{2} (x) + 3 sin (x) - 4 sin^{3} (x) = 0

令

: sin (x) = u

1 - 2 u^{2} + 3 u - 4 u^{3} = 0

1 - 2 u^{2} + 3 u - 4 u^{3} = 0 : u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

1 - 2 u^{2} + 3 u - 4 u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- 4 u^{3} - 2 u^{2} + 3 u + 1 = 0

因式分解

- 4 u^{3} - 2 u^{2} + 3 u + 1 : - (u + 1) (4 u^{2} - 2 u - 1)

- 4 u^{3} - 2 u^{2} + 3 u + 1

因式分解出通项

- 1

= - (4 u^{3} + 2 u^{2} - 3 u - 1)

分解

4 u^{3} + 2 u^{2} - 3 u - 1 : (u + 1) (4 u^{2} - 2 u - 1)

4 u^{3} + 2 u^{2} - 3 u - 1

使用有理根定理

a_{0} = 1, a_{n} = 4

a_{0}

的除数

: 1, a_{n}

的除数

: 1, 2, 4

因此，检验以下有理数:

\pm \frac{1}{1 , 2 , 4}

- \frac{1}{1}

是表达式的根，所以因式分解

u + 1

= (u + 1) \frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} - 2 u - 1

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

对

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

做除法:

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

将分子

4 u^{3} + 2 u^{2} - 3 u - 1

与除数

u + 1

的首项系数相除：

\frac{4 u ^{3}}{u} = 4 u^{2}

商 = 4 u^{2}

将

u + 1

乘以

4 u^{2} : 4 u^{3} + 4 u^{2}

将

4 u^{3} + 2 u^{2} - 3 u - 1

减去

4 u^{3} + 4 u^{2}

得到新的余数

余数 = - 2 u^{2} - 3 u - 1

因此

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

= 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

对

\frac{- 2 u ^{2} - 3 u - 1}{u + 1}

做除法:

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

将分子

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

与除数

u + 1

的首项系数相除：

\frac{- 2 u ^{2}}{u} = - 2 u

商 = - 2 u

将

u + 1

乘以

- 2 u : - 2 u^{2} - 2 u

将

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

减去

- 2 u^{2} - 2 u

得到新的余数

余数 = - u - 1

因此

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

= 4 u^{2} - 2 u + \frac{- u - 1}{u + 1}

对

\frac{- u - 1}{u + 1}

做除法:

\frac{- u - 1}{u + 1} = - 1

将分子

- u - 1

与除数

u + 1

的首项系数相除：

\frac{- u}{u} = - 1

商 = - 1

将

u + 1

乘以

- 1 : - u - 1

将

- u - 1

减去

- u - 1

得到新的余数

余数 = 0

因此

\frac{- u - 1}{u + 1} = - 1

= 4 u^{2} - 2 u - 1

= 4 u^{2} - 2 u - 1

= (u + 1) (4 u^{2} - 2 u - 1)

= - (u + 1) (4 u^{2} - 2 u - 1)

- (u + 1) (4 u^{2} - 2 u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u + 1 = 0 or 4 u^{2} - 2 u - 1 = 0

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解

4 u^{2} - 2 u - 1 = 0 : u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

4 u^{2} - 2 u - 1 = 0

使用求根公式求解

4 u^{2} - 2 u - 1 = 0

二次方程求根公式:

若

a = 4, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 4 (- 1) = 25

(- 2)^{2} - 4 \cdot 4 (- 1)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 4 \cdot 1

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数字相加:

4 + 16 = 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 ) \pm 2 5}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 4}

u = \frac{- ( - 2 ) + 2 5}{2 \cdot 4} : \frac{1 + 5}{4}

\frac{- ( - 2 ) + 2 5}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 + 2 5}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 + 2 5}{8}

分解

2 + 25 : 2 (1 + 5)

2 + 25

改写为

= 2 \cdot 1 + 25

因式分解出通项

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

约分:

2

= \frac{1 + 5}{4}

u = \frac{- ( - 2 ) - 2 5}{2 \cdot 4} : \frac{1 - 5}{4}

\frac{- ( - 2 ) - 2 5}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{2 - 2 5}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{2 - 2 5}{8}

分解

2 - 25 : 2 (1 - 5)

2 - 25

改写为

= 2 \cdot 1 - 25

因式分解出通项

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{8}

约分:

2

= \frac{1 - 5}{4}

二次方程组的解是:

u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

解为

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

u = sin (x)

代回

sin (x) = - 1, sin (x) = \frac{1 + 5}{4}, sin (x) = \frac{1 - 5}{4}

sin (x) = - 1, sin (x) = \frac{1 + 5}{4}, sin (x) = \frac{1 - 5}{4}

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) = \frac{1 + 5}{4} : x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (x) = \frac{1 + 5}{4}

使用反三角函数性质

sin (x) = \frac{1 + 5}{4}

sin (x) = \frac{1 + 5}{4}

的通解

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

x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (x) = \frac{1 - 5}{4} : x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

sin (x) = \frac{1 - 5}{4}

使用反三角函数性质

sin (x) = \frac{1 - 5}{4}

sin (x) = \frac{1 - 5}{4}

的通解

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

x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

合并所有解

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

以小数形式表示解

x = \frac{3 π}{2} + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn

作图

流行的例子

tan(x)=-(sqrt(3))/2

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

3sin(x)cos(x)=sin(x)

3 sin (x) cos (x) = sin (x)

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cos (\frac{5 x}{10}) = \frac{2}{2}

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