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受欢迎的三角函数 >

sin(x)-cos(2x)-1=0

解答

sin (x) - cos (2 x) - 1 = 0

解答

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

+1

度数

x = 51.33171 \dots^{\circ} + 36 0^{\circ} n, x = 128.66828 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin (x) - cos (2 x) - 1 = 0

使用三角恒等式改写

- 1 - cos (2 x) + sin (x)

使用倍角公式:

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

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

化简

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

数字相减:

- 1 - 1 = - 2

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

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

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

用替代法求解

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

令

: sin (x) = u

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

- 2 + u + 2 u^{2} = 0 : u = \frac{- 1 + 17}{4}, u = \frac{- 1 - 17}{4}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 \cdot 2 (- 2) = 17

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 1 + 17}{2 \cdot 2}, u_{2} = \frac{- 1 - 17}{2 \cdot 2}

u = \frac{- 1 + 17}{2 \cdot 2} : \frac{- 1 + 17}{4}

\frac{- 1 + 17}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 + 17}{4}

u = \frac{- 1 - 17}{2 \cdot 2} : \frac{- 1 - 17}{4}

\frac{- 1 - 17}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 - 17}{4}

二次方程组的解是:

u = \frac{- 1 + 17}{4}, u = \frac{- 1 - 17}{4}

u = sin (x)

代回

sin (x) = \frac{- 1 + 17}{4}, sin (x) = \frac{- 1 - 17}{4}

sin (x) = \frac{- 1 + 17}{4}, sin (x) = \frac{- 1 - 17}{4}

sin (x) = \frac{- 1 + 17}{4} : x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{- 1 + 17}{4}

使用反三角函数性质

sin (x) = \frac{- 1 + 17}{4}

sin (x) = \frac{- 1 + 17}{4}

的通解

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

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{- 1 - 17}{4} :

无解

sin (x) = \frac{- 1 - 17}{4}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

以小数形式表示解

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

作图

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