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受欢迎的三角函数 >

4cos(2θ)-10cos(θ)+14=7,0<= θ<360

解答

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

解答

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

+1

弧度

θ = 0.72273 \dots, θ = 2 π - 0.72273 \dots, θ = \frac{π}{3}, θ = \frac{5 π}{3}

求解步骤

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

两边减去

7

4 cos (2 θ) - 10 cos (θ) + 7 = 0

使用三角恒等式改写

7 - 10 cos (θ) + 4 cos (2 θ)

使用倍角公式:

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

= 7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1)

化简

7 - 10 cos (θ) + 4 (2 cos^{2} (θ) - 1) : 8 cos^{2} (θ) - 10 cos (θ) + 3

7 - 10 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

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

= 7 - 10 cos (θ) + 8 cos^{2} (θ) - 4

化简

7 - 10 cos (θ) + 8 cos^{2} (θ) - 4 : 8 cos^{2} (θ) - 10 cos (θ) + 3

7 - 10 cos (θ) + 8 cos^{2} (θ) - 4

对同类项分组

= - 10 cos (θ) + 8 cos^{2} (θ) + 7 - 4

数字相加/相减:

7 - 4 = 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

= 8 cos^{2} (θ) - 10 cos (θ) + 3

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

用替代法求解

3 - 10 cos (θ) + 8 cos^{2} (θ) = 0

令

: cos (θ) = u

3 - 10 u + 8 u^{2} = 0

3 - 10 u + 8 u^{2} = 0 : u = \frac{3}{4}, u = \frac{1}{2}

3 - 10 u + 8 u^{2} = 0

改写成标准形式

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

8 u^{2} - 10 u + 3 = 0

使用求根公式求解

8 u^{2} - 10 u + 3 = 0

二次方程求根公式:

若

a = 8, b = - 10, c = 3

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 8 \cdot 3}{2 \cdot 8}

(- 10)^{2} - 4 \cdot 8 \cdot 3 = 2

(- 10)^{2} - 4 \cdot 8 \cdot 3

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 8 \cdot 3

数字相乘:

4 \cdot 8 \cdot 3 = 96

= 1 0^{2} - 96

1 0^{2} = 100

= 100 - 96

数字相减:

100 - 96 = 4

= 4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 10 ) \pm 2}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 10 ) + 2}{2 \cdot 8}, u_{2} = \frac{- ( - 10 ) - 2}{2 \cdot 8}

u = \frac{- ( - 10 ) + 2}{2 \cdot 8} : \frac{3}{4}

\frac{- ( - 10 ) + 2}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{10 + 2}{2 \cdot 8}

数字相加:

10 + 2 = 12

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

数字相乘:

2 \cdot 8 = 16

= \frac{12}{16}

约分:

4

= \frac{3}{4}

u = \frac{- ( - 10 ) - 2}{2 \cdot 8} : \frac{1}{2}

\frac{- ( - 10 ) - 2}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{10 - 2}{2 \cdot 8}

数字相减:

10 - 2 = 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

二次方程组的解是:

u = \frac{3}{4}, u = \frac{1}{2}

u = cos (θ)

代回

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{2}

cos (θ) = \frac{3}{4}, cos (θ) = \frac{1}{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{1}{2}, 0 \leq θ < 36 0^{\circ} : θ = 6 0^{\circ}, θ = 30 0^{\circ}

cos (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ}

cos (θ) = \frac{1}{2}

的通解

cos (x)

周期表（周期为

36 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

在

0 \leq θ < 36 0^{\circ}

范围内的解

θ = 6 0^{\circ}, θ = 30 0^{\circ}

合并所有解

θ = arccos (\frac{3}{4}), θ = 36 0^{\circ} - arccos (\frac{3}{4}), θ = 6 0^{\circ}, θ = 30 0^{\circ}

以小数形式表示解

θ = 0.72273 \dots, θ = 36 0^{\circ} - 0.72273 \dots, θ = 6 0^{\circ}, θ = 30 0^{\circ}

作图

流行的例子

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tan(2x-pi/6)= 1/(3^{1/2)}

sin^2(θ)=cos(θ)+cos^2(θ)