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受欢迎的三角函数 >

3sin(x)-4sin^3(x)=1-2sin^2(x)

解答

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

解答

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

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 1 8^{\circ} + 36 0^{\circ} n, x = 16 2^{\circ} + 36 0^{\circ} n, x = - 5 4^{\circ} + 36 0^{\circ} n, x = 23 4^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: sin (x) = u

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

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

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

将

2 u^{2}

para o lado esquerdo

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

两边加上

2 u^{2}

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

化简

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

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

将

1

para o lado esquerdo

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

两边减去

1

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

化简

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

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

两边加上

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

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}

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}

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

作图

流行的例子

sin^2(a)=((2tan(a)))/((1+tan^2(a)))

sin^{2} (a) = \frac{( 2 tan ( a ) )}{( 1 + tan ^{2} ( a ) )}

(tan(a))/2 = 2/11

\frac{tan ( a )}{2} = \frac{2}{11}

sin^2(x)-cos(x)= 1/2

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

cos(x)[3sin(x)-2]=0

cos (x) [3 sin (x) - 2] = 0

sin^3(x)+sin(x)-4=0

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