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受欢迎的三角函数 >

-2cos^2(4x)+2sin(4x)-1=0

解答

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

解答

x = \frac{0.96645 \dots}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{0.96645 \dots}{4} + \frac{πn}{2}

+1

度数

x = 13.84342 \dots^{\circ} + 9 0^{\circ} n, x = 31.15657 \dots^{\circ} + 9 0^{\circ} n

求解步骤

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

使用三角恒等式改写

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (4 x)

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

使用加减运算法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

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

数字相减:

- 1 - 2 = - 3

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

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

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

用替代法求解

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

令

: sin (4 x) = u

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

- 3 + 2 u + 2 u^{2} = 0 : u = \frac{- 1 + 7}{2}, u = - \frac{1 + 7}{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = 2, c = - 3

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

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

2^{2} - 4 \cdot 2 (- 3) = 27

2^{2} - 4 \cdot 2 (- 3)

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 2^{2} + 24

2^{2} = 4

= 4 + 24

数字相加:

4 + 24 = 28

= 28

28

质因数分解:

2^{2} \cdot 7

28

28

除以

228 = 14 \cdot 2

= 2 \cdot 14

14

除以

214 = 7 \cdot 2

= 2 \cdot 2 \cdot 7

2, 7

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 7

= 2^{2} \cdot 7

= 2^{2} \cdot 7

使用根式运算法则:

n ab = n a n b

= 7 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 27

u_{1, 2} = \frac{- 2 \pm 2 7}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 2 + 2 7}{2 \cdot 2}, u_{2} = \frac{- 2 - 2 7}{2 \cdot 2}

u = \frac{- 2 + 2 7}{2 \cdot 2} : \frac{- 1 + 7}{2}

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 2 7}{4}

分解

- 2 + 27 : 2 (- 1 + 7)

- 2 + 27

改写为

= - 2 \cdot 1 + 27

因式分解出通项

2

= 2 (- 1 + 7)

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

约分:

2

= \frac{- 1 + 7}{2}

u = \frac{- 2 - 2 7}{2 \cdot 2} : - \frac{1 + 7}{2}

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 2 7}{4}

分解

- 2 - 27 : - 2 (1 + 7)

- 2 - 27

改写为

= - 2 \cdot 1 - 27

因式分解出通项

2

= - 2 (1 + 7)

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

约分:

2

= - \frac{1 + 7}{2}

二次方程组的解是:

u = \frac{- 1 + 7}{2}, u = - \frac{1 + 7}{2}

u = sin (4 x)

代回

sin (4 x) = \frac{- 1 + 7}{2}, sin (4 x) = - \frac{1 + 7}{2}

sin (4 x) = \frac{- 1 + 7}{2}, sin (4 x) = - \frac{1 + 7}{2}

sin (4 x) = \frac{- 1 + 7}{2} : x = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

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

使用反三角函数性质

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

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

的通解

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

4 x = arcsin (\frac{- 1 + 7}{2}) + 2 πn, 4 x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

4 x = arcsin (\frac{- 1 + 7}{2}) + 2 πn, 4 x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

解

4 x = arcsin (\frac{- 1 + 7}{2}) + 2 πn : x = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

4 x = arcsin (\frac{- 1 + 7}{2}) + 2 πn

两边除以

4

4 x = arcsin (\frac{- 1 + 7}{2}) + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{2 πn}{4}

化简

x = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

x = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

解

4 x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn : x = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

4 x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

两边除以

4

4 x = π - arcsin (\frac{- 1 + 7}{2}) + 2 πn

两边除以

4

\frac{4 x}{4} = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{2 πn}{4}

化简

x = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

x = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

x = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

sin (4 x) = - \frac{1 + 7}{2} :

无解

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

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{arcsin ( \frac{- 1 + 7}{2} )}{4} + \frac{πn}{2}

以小数形式表示解

x = \frac{0.96645 \dots}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{0.96645 \dots}{4} + \frac{πn}{2}

作图

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