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受欢迎的三角函数 >

4cos(2θ)=cos^2(θ)-2

解答

4 cos (2 θ) = cos^{2} (θ) - 2

解答

θ = 1.00685 \dots + 2 πn, θ = 2 π - 1.00685 \dots + 2 πn, θ = 2.13473 \dots + 2 πn, θ = - 2.13473 \dots + 2 πn

+1

度数

θ = 57.68846 \dots^{\circ} + 36 0^{\circ} n, θ = 302.31153 \dots^{\circ} + 36 0^{\circ} n, θ = 122.31153 \dots^{\circ} + 36 0^{\circ} n, θ = - 122.31153 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (2 θ) = cos^{2} (θ) - 2

两边减去

cos^{2} (θ) - 2

4 cos (2 θ) - cos^{2} (θ) + 2 = 0

使用三角恒等式改写

2 - cos^{2} (θ) + 4 cos (2 θ)

使用倍角公式:

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

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

化简

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

2 - cos^{2} (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

= 2 - cos^{2} (θ) + 8 cos^{2} (θ) - 4

化简

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

2 - cos^{2} (θ) + 8 cos^{2} (θ) - 4

同类项相加:

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

= 2 + 7 cos^{2} (θ) - 4

对同类项分组

= 7 cos^{2} (θ) + 2 - 4

数字相加/相减:

2 - 4 = - 2

= 7 cos^{2} (θ) - 2

= 7 cos^{2} (θ) - 2

= 7 cos^{2} (θ) - 2

- 2 + 7 cos^{2} (θ) = 0

用替代法求解

- 2 + 7 cos^{2} (θ) = 0

令

: cos (θ) = u

- 2 + 7 u^{2} = 0

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

- 2 + 7 u^{2} = 0

将

2

到右边

- 2 + 7 u^{2} = 0

两边加上

2

- 2 + 7 u^{2} + 2 = 0 + 2

化简

7 u^{2} = 2

7 u^{2} = 2

两边除以

7

7 u^{2} = 2

两边除以

7

\frac{7 u ^{2}}{7} = \frac{2}{7}

化简

u^{2} = \frac{2}{7}

u^{2} = \frac{2}{7}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

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

u = cos (θ)

代回

cos (θ) = \frac{2}{7}, cos (θ) = - \frac{2}{7}

cos (θ) = \frac{2}{7}, cos (θ) = - \frac{2}{7}

cos (θ) = \frac{2}{7} : θ = arccos (\frac{2}{7}) + 2 πn, θ = 2 π - arccos (\frac{2}{7}) + 2 πn

cos (θ) = \frac{2}{7}

使用反三角函数性质

cos (θ) = \frac{2}{7}

cos (θ) = \frac{2}{7}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{2}{7}) + 2 πn, θ = 2 π - arccos (\frac{2}{7}) + 2 πn

θ = arccos (\frac{2}{7}) + 2 πn, θ = 2 π - arccos (\frac{2}{7}) + 2 πn

cos (θ) = - \frac{2}{7} : θ = arccos (- \frac{2}{7}) + 2 πn, θ = - arccos (- \frac{2}{7}) + 2 πn

cos (θ) = - \frac{2}{7}

使用反三角函数性质

cos (θ) = - \frac{2}{7}

cos (θ) = - \frac{2}{7}

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

θ = arccos (- \frac{2}{7}) + 2 πn, θ = - arccos (- \frac{2}{7}) + 2 πn

θ = arccos (- \frac{2}{7}) + 2 πn, θ = - arccos (- \frac{2}{7}) + 2 πn

合并所有解

θ = arccos (\frac{2}{7}) + 2 πn, θ = 2 π - arccos (\frac{2}{7}) + 2 πn, θ = arccos (- \frac{2}{7}) + 2 πn, θ = - arccos (- \frac{2}{7}) + 2 πn

以小数形式表示解

θ = 1.00685 \dots + 2 πn, θ = 2 π - 1.00685 \dots + 2 πn, θ = 2.13473 \dots + 2 πn, θ = - 2.13473 \dots + 2 πn

作图

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