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受欢迎的三角函数 >

cos^2(x)-5cos(x)-1=0

解答

cos^{2} (x) - 5 cos (x) - 1 = 0

解答

x = 1.76458 \dots + 2 πn, x = - 1.76458 \dots + 2 πn

+1

度数

x = 101.10352 \dots^{\circ} + 36 0^{\circ} n, x = - 101.10352 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (x) - 5 cos (x) - 1 = 0

用替代法求解

cos^{2} (x) - 5 cos (x) - 1 = 0

令

: cos (x) = u

u^{2} - 5 u - 1 = 0

u^{2} - 5 u - 1 = 0 : u = \frac{5 + 29}{2}, u = \frac{5 - 29}{2}

u^{2} - 5 u - 1 = 0

使用求根公式求解

u^{2} - 5 u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 5, c = - 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot (- 1) = 29

(- 5)^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= (- 5)^{2} + 4 \cdot 1 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 5^{2} + 4

5^{2} = 25

= 25 + 4

数字相加:

25 + 4 = 29

= 29

u_{1, 2} = \frac{- ( - 5 ) \pm 29}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 29}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 29}{2 \cdot 1}

u = \frac{- ( - 5 ) + 29}{2 \cdot 1} : \frac{5 + 29}{2}

\frac{- ( - 5 ) + 29}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 + 29}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 + 29}{2}

u = \frac{- ( - 5 ) - 29}{2 \cdot 1} : \frac{5 - 29}{2}

\frac{- ( - 5 ) - 29}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{5 - 29}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{5 - 29}{2}

二次方程组的解是:

u = \frac{5 + 29}{2}, u = \frac{5 - 29}{2}

u = cos (x)

代回

cos (x) = \frac{5 + 29}{2}, cos (x) = \frac{5 - 29}{2}

cos (x) = \frac{5 + 29}{2}, cos (x) = \frac{5 - 29}{2}

cos (x) = \frac{5 + 29}{2} :

无解

cos (x) = \frac{5 + 29}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{5 - 29}{2} : x = arccos (\frac{5 - 29}{2}) + 2 πn, x = - arccos (\frac{5 - 29}{2}) + 2 πn

cos (x) = \frac{5 - 29}{2}

使用反三角函数性质

cos (x) = \frac{5 - 29}{2}

cos (x) = \frac{5 - 29}{2}

的通解

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

x = arccos (\frac{5 - 29}{2}) + 2 πn, x = - arccos (\frac{5 - 29}{2}) + 2 πn

x = arccos (\frac{5 - 29}{2}) + 2 πn, x = - arccos (\frac{5 - 29}{2}) + 2 πn

合并所有解

x = arccos (\frac{5 - 29}{2}) + 2 πn, x = - arccos (\frac{5 - 29}{2}) + 2 πn

以小数形式表示解

x = 1.76458 \dots + 2 πn, x = - 1.76458 \dots + 2 πn

作图

流行的例子

cos(2θ)=-cos(θ)

cos (2 θ) = - cos (θ)

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

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

2cos^2(θ)=1-cos(θ)

2 cos^{2} (θ) = 1 - cos (θ)

-5cos^2(θ)=12sin(θ)+4

- 5 cos^{2} (θ) = 12 sin (θ) + 4

cos^2(t)-2sin(t)-1=0

cos^{2} (t) - 2 sin (t) - 1 = 0