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受欢迎的三角函数 >

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

解答

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

解答

x = - 0.40105 \dots + 2 πn, x = π + 0.40105 \dots + 2 πn, x = 0.69500 \dots + 2 πn, x = π - 0.69500 \dots + 2 πn

+1

度数

x = - 22.97865 \dots^{\circ} + 36 0^{\circ} n, x = 202.97865 \dots^{\circ} + 36 0^{\circ} n, x = 39.82077 \dots^{\circ} + 36 0^{\circ} n, x = 140.17922 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

1

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

化简

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

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

数字相乘:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

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

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

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

化简

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

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

对同类项分组

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

1 + u - 4 u^{2} = 0 : u = - \frac{- 1 + 17}{8}, u = \frac{1 + 17}{8}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 4) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 1 + 17}{2 ( - 4 )}, u_{2} = \frac{- 1 - 17}{2 ( - 4 )}

u = \frac{- 1 + 17}{2 ( - 4 )} : - \frac{- 1 + 17}{8}

\frac{- 1 + 17}{2 ( - 4 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 + 17}{- 8}

使用分式法则:

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

= - \frac{- 1 + 17}{8}

u = \frac{- 1 - 17}{2 ( - 4 )} : \frac{1 + 17}{8}

\frac{- 1 - 17}{2 ( - 4 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 - 17}{- 8}

使用分式法则:

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

- 1 - 17 = - (1 + 17)

= \frac{1 + 17}{8}

二次方程组的解是:

u = - \frac{- 1 + 17}{8}, u = \frac{1 + 17}{8}

u = sin (x)

代回

sin (x) = - \frac{- 1 + 17}{8}, sin (x) = \frac{1 + 17}{8}

sin (x) = - \frac{- 1 + 17}{8}, sin (x) = \frac{1 + 17}{8}

sin (x) = - \frac{- 1 + 17}{8} : x = arcsin (- \frac{- 1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{8}) + 2 πn

sin (x) = - \frac{- 1 + 17}{8}

使用反三角函数性质

sin (x) = - \frac{- 1 + 17}{8}

sin (x) = - \frac{- 1 + 17}{8}

的通解

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

x = arcsin (- \frac{- 1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{8}) + 2 πn

x = arcsin (- \frac{- 1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{8}) + 2 πn

sin (x) = \frac{1 + 17}{8} : x = arcsin (\frac{1 + 17}{8}) + 2 πn, x = π - arcsin (\frac{1 + 17}{8}) + 2 πn

sin (x) = \frac{1 + 17}{8}

使用反三角函数性质

sin (x) = \frac{1 + 17}{8}

sin (x) = \frac{1 + 17}{8}

的通解

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

x = arcsin (\frac{1 + 17}{8}) + 2 πn, x = π - arcsin (\frac{1 + 17}{8}) + 2 πn

x = arcsin (\frac{1 + 17}{8}) + 2 πn, x = π - arcsin (\frac{1 + 17}{8}) + 2 πn

合并所有解

x = arcsin (- \frac{- 1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{8}) + 2 πn, x = arcsin (\frac{1 + 17}{8}) + 2 πn, x = π - arcsin (\frac{1 + 17}{8}) + 2 πn

以小数形式表示解

x = - 0.40105 \dots + 2 πn, x = π + 0.40105 \dots + 2 πn, x = 0.69500 \dots + 2 πn, x = π - 0.69500 \dots + 2 πn

作图

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