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1/(cos^2(θ/2))=12cos(θ)

解答

\frac{1}{cos ^{2} ( \frac{θ}{2} )} = 12 cos (θ)

解答

θ = 1.42478 \dots + 2 πn, θ = 2 π - 1.42478 \dots + 2 πn

+1

度数

θ = 81.63392 \dots^{\circ} + 36 0^{\circ} n, θ = 278.36607 \dots^{\circ} + 36 0^{\circ} n

求解步骤

\frac{1}{cos ^{2} ( \frac{θ}{2} )} = 12 cos (θ)

两边减去

12 cos (θ)

\frac{1}{cos ^{2} ( \frac{θ}{2} )} - 12 cos (θ) = 0

化简

\frac{1}{cos ^{2} ( \frac{θ}{2} )} - 12 cos (θ) : \frac{1 - 12 cos ^{2} ( \frac{θ}{2} ) cos ( θ )}{cos ^{2} ( \frac{θ}{2} )}

\frac{1}{cos ^{2} ( \frac{θ}{2} )} - 12 cos (θ)

将项转换为分式:

12 cos (θ) = \frac{12 cos ( θ ) cos ^{2} ( \frac{θ}{2} )}{cos ^{2} ( \frac{θ}{2} )}

= \frac{1}{cos ^{2} ( \frac{θ}{2} )} - \frac{12 cos ( θ ) cos ^{2} ( \frac{θ}{2} )}{cos ^{2} ( \frac{θ}{2} )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - 12 cos ( θ ) cos ^{2} ( \frac{θ}{2} )}{cos ^{2} ( \frac{θ}{2} )}

\frac{1 - 12 cos ^{2} ( \frac{θ}{2} ) cos ( θ )}{cos ^{2} ( \frac{θ}{2} )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 - 12 cos^{2} (\frac{θ}{2}) cos (θ) = 0

使用三角恒等式改写

1 - 12 cos^{2} (\frac{θ}{2}) cos (θ)

利用以下特性:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 2 cos^{2} (θ) - 1

交换两边

2 cos^{2} (θ) - 1 = cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 + cos (2 θ)

两边除以

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

= 1 - 12 \cdot \frac{1 + cos ( 2 \cdot \frac{θ}{2} )}{2} cos (θ)

12 \cdot \frac{1 + cos ( 2 \cdot \frac{θ}{2} )}{2} cos (θ) = 6 cos (θ) (cos (θ) + 1)

12 \cdot \frac{1 + cos ( 2 \cdot \frac{θ}{2} )}{2} cos (θ)

\frac{1 + cos ( 2 \cdot \frac{θ}{2} )}{2} = \frac{1 + cos ( θ )}{2}

\frac{1 + cos ( 2 \cdot \frac{θ}{2} )}{2}

乘

2 \cdot \frac{θ}{2} : θ

2 \cdot \frac{θ}{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{θ \cdot 2}{2}

约分:

2

= θ

= \frac{1 + cos ( θ )}{2}

= 12 \cdot \frac{cos ( θ ) + 1}{2} cos (θ)

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 1 + cos ( θ ) ) \cdot 12 cos ( θ )}{2}

数字相除:

\frac{12}{2} = 6

= 6 cos (θ) (cos (θ) + 1)

= 1 - 6 cos (θ) (cos (θ) + 1)

1 - (1 + cos (θ)) \cdot 6 cos (θ) = 0

用替代法求解

1 - (1 + cos (θ)) \cdot 6 cos (θ) = 0

令

: cos (θ) = u

1 - (1 + u) \cdot 6 u = 0

1 - (1 + u) \cdot 6 u = 0 : u = - \frac{3 + 15}{6}, u = \frac{15 - 3}{6}

1 - (1 + u) \cdot 6 u = 0

展开

1 - (1 + u) \cdot 6 u : 1 - 6 u - 6 u^{2}

1 - (1 + u) \cdot 6 u

= 1 - 6 u (1 + u)

乘开

- 6 u (1 + u) : - 6 u - 6 u^{2}

- 6 u (1 + u)

使用分配律:

a (b + c) = ab + a c

a = - 6 u, b = 1, c = u

= - 6 u \cdot 1 + (- 6 u) u

使用加减运算法则

+ (- a) = - a

= - 6 \cdot 1 \cdot u - 6 uu

化简

- 6 \cdot 1 \cdot u - 6 uu : - 6 u - 6 u^{2}

- 6 \cdot 1 \cdot u - 6 uu

6 \cdot 1 \cdot u = 6 u

6 \cdot 1 \cdot u

数字相乘:

6 \cdot 1 = 6

= 6 u

6 uu = 6 u^{2}

6 uu

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= 6 u^{1 + 1}

数字相加:

1 + 1 = 2

= 6 u^{2}

= - 6 u - 6 u^{2}

= - 6 u - 6 u^{2}

= 1 - 6 u - 6 u^{2}

1 - 6 u - 6 u^{2} = 0

改写成标准形式

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

- 6 u^{2} - 6 u + 1 = 0

使用求根公式求解

- 6 u^{2} - 6 u + 1 = 0

二次方程求根公式:

若

a = - 6, b = - 6, c = 1

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

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

(- 6)^{2} - 4 (- 6) \cdot 1 = 215

(- 6)^{2} - 4 (- 6) \cdot 1

使用法则

- (- a) = a

= (- 6)^{2} + 4 \cdot 6 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 6)^{2} = 6^{2}

= 6^{2} + 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 6^{2} + 24

6^{2} = 36

= 36 + 24

数字相加:

36 + 24 = 60

= 60

60

质因数分解:

2^{2} \cdot 3 \cdot 5

60

60

除以

260 = 30 \cdot 2

= 2 \cdot 30

30

除以

230 = 15 \cdot 2

= 2 \cdot 2 \cdot 15

15

除以

315 = 5 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 5

2, 3, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3 \cdot 5

= 2^{2} \cdot 3 \cdot 5

= 2^{2} \cdot 3 \cdot 5

使用根式运算法则:

= 2^{2} 3 \cdot 5

使用根式运算法则:

2^{2} = 2

= 2 3 \cdot 5

整理后得

= 215

u_{1, 2} = \frac{- ( - 6 ) \pm 2 15}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- ( - 6 ) + 2 15}{2 ( - 6 )}, u_{2} = \frac{- ( - 6 ) - 2 15}{2 ( - 6 )}

u = \frac{- ( - 6 ) + 2 15}{2 ( - 6 )} : - \frac{3 + 15}{6}

\frac{- ( - 6 ) + 2 15}{2 ( - 6 )}

去除括号:

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

= \frac{6 + 2 15}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6 + 2 15}{- 12}

使用分式法则:

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

= - \frac{6 + 2 15}{12}

消掉

\frac{6 + 2 15}{12} : \frac{3 + 15}{6}

\frac{6 + 2 15}{12}

分解

6 + 215 : 2 (3 + 15)

6 + 215

改写为

= 2 \cdot 3 + 215

因式分解出通项

2

= 2 (3 + 15)

= \frac{2 ( 3 + 15 )}{12}

约分:

2

= \frac{3 + 15}{6}

= - \frac{3 + 15}{6}

u = \frac{- ( - 6 ) - 2 15}{2 ( - 6 )} : \frac{15 - 3}{6}

\frac{- ( - 6 ) - 2 15}{2 ( - 6 )}

去除括号:

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

= \frac{6 - 2 15}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6 - 2 15}{- 12}

使用分式法则:

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

6 - 215 = - (215 - 6)

= \frac{2 15 - 6}{12}

分解

215 - 6 : 2 (15 - 3)

215 - 6

改写为

= 215 - 2 \cdot 3

因式分解出通项

2

= 2 (15 - 3)

= \frac{2 ( 15 - 3 )}{12}

约分:

2

= \frac{15 - 3}{6}

二次方程组的解是:

u = - \frac{3 + 15}{6}, u = \frac{15 - 3}{6}

u = cos (θ)

代回

cos (θ) = - \frac{3 + 15}{6}, cos (θ) = \frac{15 - 3}{6}

cos (θ) = - \frac{3 + 15}{6}, cos (θ) = \frac{15 - 3}{6}

cos (θ) = - \frac{3 + 15}{6} :

无解

cos (θ) = - \frac{3 + 15}{6}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = \frac{15 - 3}{6} : θ = arccos (\frac{15 - 3}{6}) + 2 πn, θ = 2 π - arccos (\frac{15 - 3}{6}) + 2 πn

cos (θ) = \frac{15 - 3}{6}

使用反三角函数性质

cos (θ) = \frac{15 - 3}{6}

cos (θ) = \frac{15 - 3}{6}

的通解

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

θ = arccos (\frac{15 - 3}{6}) + 2 πn, θ = 2 π - arccos (\frac{15 - 3}{6}) + 2 πn

θ = arccos (\frac{15 - 3}{6}) + 2 πn, θ = 2 π - arccos (\frac{15 - 3}{6}) + 2 πn

合并所有解

θ = arccos (\frac{15 - 3}{6}) + 2 πn, θ = 2 π - arccos (\frac{15 - 3}{6}) + 2 πn

以小数形式表示解

θ = 1.42478 \dots + 2 πn, θ = 2 π - 1.42478 \dots + 2 πn

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