zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

2cos^2(θ)-5cos(θ)+1=0

解答

2 cos^{2} (θ) - 5 cos (θ) + 1 = 0

解答

θ = 1.34977 \dots + 2 πn, θ = 2 π - 1.34977 \dots + 2 πn

+1

度数

θ = 77.33656 \dots^{\circ} + 36 0^{\circ} n, θ = 282.66343 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (θ) - 5 cos (θ) + 1 = 0

用替代法求解

2 cos^{2} (θ) - 5 cos (θ) + 1 = 0

令

: cos (θ) = u

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

2 u^{2} - 5 u + 1 = 0 : u = \frac{5 + 17}{4}, u = \frac{5 - 17}{4}

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

使用求根公式求解

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

二次方程求根公式:

若

a = 2, b = - 5, c = 1

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

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

(- 5)^{2} - 4 \cdot 2 \cdot 1 = 17

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

数字相减:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- ( - 5 ) \pm 17}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 17}{2 \cdot 2}, u_{2} = \frac{- ( - 5 ) - 17}{2 \cdot 2}

u = \frac{- ( - 5 ) + 17}{2 \cdot 2} : \frac{5 + 17}{4}

\frac{- ( - 5 ) + 17}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 + 17}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{5 + 17}{4}

u = \frac{- ( - 5 ) - 17}{2 \cdot 2} : \frac{5 - 17}{4}

\frac{- ( - 5 ) - 17}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{5 - 17}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{5 - 17}{4}

二次方程组的解是:

u = \frac{5 + 17}{4}, u = \frac{5 - 17}{4}

u = cos (θ)

代回

cos (θ) = \frac{5 + 17}{4}, cos (θ) = \frac{5 - 17}{4}

cos (θ) = \frac{5 + 17}{4}, cos (θ) = \frac{5 - 17}{4}

cos (θ) = \frac{5 + 17}{4} :

无解

cos (θ) = \frac{5 + 17}{4}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = \frac{5 - 17}{4} : θ = arccos (\frac{5 - 17}{4}) + 2 πn, θ = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

cos (θ) = \frac{5 - 17}{4}

使用反三角函数性质

cos (θ) = \frac{5 - 17}{4}

cos (θ) = \frac{5 - 17}{4}

的通解

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

θ = arccos (\frac{5 - 17}{4}) + 2 πn, θ = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

θ = arccos (\frac{5 - 17}{4}) + 2 πn, θ = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

合并所有解

θ = arccos (\frac{5 - 17}{4}) + 2 πn, θ = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

以小数形式表示解

θ = 1.34977 \dots + 2 πn, θ = 2 π - 1.34977 \dots + 2 πn

作图

流行的例子

8sin(θ)cos(θ)=3

8 sin (θ) cos (θ) = 3

tan (x) = \frac{7}{9}

cos (8 x) = - 1

-3cos^2(x)+2cos(x)+1=0

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

sin^2(x)-2=cos(2x)

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