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受欢迎的三角函数 >

-4cos(2θ)-19=-26cos(θ),0<= θ<360

解答

- 4 cos (2 θ) - 19 = - 26 cos (θ), 0 \leq θ < 36 0^{\circ}

解答

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots

+1

弧度

θ = 0.72273 \dots, θ = 2 π - 0.72273 \dots

求解步骤

- 4 cos (2 θ) - 19 = - 26 cos (θ), 0 \leq θ < 36 0^{\circ}

两边减去

- 26 cos (θ)

- 4 cos (2 θ) - 19 + 26 cos (θ) = 0

使用三角恒等式改写

- 19 + 26 cos (θ) - 4 cos (2 θ)

使用倍角公式:

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

= - 19 + 26 cos (θ) - 4 (2 cos^{2} (θ) - 1)

化简

- 19 + 26 cos (θ) - 4 (2 cos^{2} (θ) - 1) : 26 cos (θ) - 8 cos^{2} (θ) - 15

- 19 + 26 cos (θ) - 4 (2 cos^{2} (θ) - 1)

乘开

- 4 (2 cos^{2} (θ) - 1) : - 8 cos^{2} (θ) + 4

- 4 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = - 4, b = 2 cos^{2} (θ), c = 1

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

使用加减运算法则

- (- a) = a

= - 4 \cdot 2 cos^{2} (θ) + 4 \cdot 1

化简

- 4 \cdot 2 cos^{2} (θ) + 4 \cdot 1 : - 8 cos^{2} (θ) + 4

- 4 \cdot 2 cos^{2} (θ) + 4 \cdot 1

数字相乘:

4 \cdot 2 = 8

= - 8 cos^{2} (θ) + 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= - 8 cos^{2} (θ) + 4

= - 8 cos^{2} (θ) + 4

= - 19 + 26 cos (θ) - 8 cos^{2} (θ) + 4

化简

- 19 + 26 cos (θ) - 8 cos^{2} (θ) + 4 : 26 cos (θ) - 8 cos^{2} (θ) - 15

- 19 + 26 cos (θ) - 8 cos^{2} (θ) + 4

对同类项分组

= 26 cos (θ) - 8 cos^{2} (θ) - 19 + 4

数字相加/相减:

- 19 + 4 = - 15

= 26 cos (θ) - 8 cos^{2} (θ) - 15

= 26 cos (θ) - 8 cos^{2} (θ) - 15

= 26 cos (θ) - 8 cos^{2} (θ) - 15

- 15 + 26 cos (θ) - 8 cos^{2} (θ) = 0

用替代法求解

- 15 + 26 cos (θ) - 8 cos^{2} (θ) = 0

令

: cos (θ) = u

- 15 + 26 u - 8 u^{2} = 0

- 15 + 26 u - 8 u^{2} = 0 : u = \frac{3}{4}, u = \frac{5}{2}

- 15 + 26 u - 8 u^{2} = 0

改写成标准形式

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

- 8 u^{2} + 26 u - 15 = 0

使用求根公式求解

- 8 u^{2} + 26 u - 15 = 0

二次方程求根公式:

若

a = - 8, b = 26, c = - 15

u_{1, 2} = \frac{- 26 \pm 2 6 ^{2} - 4 ( - 8 ) ( - 15 )}{2 ( - 8 )}

u_{1, 2} = \frac{- 26 \pm 2 6 ^{2} - 4 ( - 8 ) ( - 15 )}{2 ( - 8 )}

2 6^{2} - 4 (- 8) (- 15) = 14

2 6^{2} - 4 (- 8) (- 15)

使用法则

- (- a) = a

= 2 6^{2} - 4 \cdot 8 \cdot 15

数字相乘:

4 \cdot 8 \cdot 15 = 480

= 2 6^{2} - 480

2 6^{2} = 676

= 676 - 480

数字相减:

676 - 480 = 196

= 196

因式分解数字:

196 = 1 4^{2}

= 1 4^{2}

使用根式运算法则:

n a^{n} = a

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- 26 \pm 14}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- 26 + 14}{2 ( - 8 )}, u_{2} = \frac{- 26 - 14}{2 ( - 8 )}

u = \frac{- 26 + 14}{2 ( - 8 )} : \frac{3}{4}

\frac{- 26 + 14}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 26 + 14}{- 2 \cdot 8}

数字相加/相减:

- 26 + 14 = - 12

= \frac{- 12}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 12}{- 16}

使用分式法则:

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

= \frac{12}{16}

约分:

4

= \frac{3}{4}

u = \frac{- 26 - 14}{2 ( - 8 )} : \frac{5}{2}

\frac{- 26 - 14}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 26 - 14}{- 2 \cdot 8}

数字相减:

- 26 - 14 = - 40

= \frac{- 40}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 40}{- 16}

使用分式法则:

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

= \frac{40}{16}

约分:

8

= \frac{5}{2}

二次方程组的解是:

u = \frac{3}{4}, u = \frac{5}{2}

u = cos (θ)

代回

cos (θ) = \frac{3}{4}, cos (θ) = \frac{5}{2}

cos (θ) = \frac{3}{4}, cos (θ) = \frac{5}{2}

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ} : θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{3}{4}, 0 \leq θ < 36 0^{\circ}

使用反三角函数性质

cos (θ) = \frac{3}{4}

cos (θ) = \frac{3}{4}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{3}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

θ = arccos (\frac{3}{4}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{4}) + 36 0^{\circ} n

在

0 \leq θ < 36 0^{\circ}

范围内的解

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

cos (θ) = \frac{5}{2}, 0 \leq θ < 36 0^{\circ} :

无解

cos (θ) = \frac{5}{2}, 0 \leq θ < 36 0^{\circ}

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4})

以小数形式表示解

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots

作图

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