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 >

4cos^2(x)+cos(x)-2=0

解答

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

解答

x = 0.93592 \dots + 2 πn, x = 2 π - 0.93592 \dots + 2 πn, x = 2.57376 \dots + 2 πn, x = - 2.57376 \dots + 2 πn

+1

度数

x = 53.62480 \dots^{\circ} + 36 0^{\circ} n, x = 306.37519 \dots^{\circ} + 36 0^{\circ} n, x = 147.46577 \dots^{\circ} + 36 0^{\circ} n, x = - 147.46577 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

4 u^{2} + u - 2 = 0 : u = \frac{- 1 + 33}{8}, u = \frac{- 1 - 33}{8}

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

使用求根公式求解

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

二次方程求根公式:

若

a = 4, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 4 (- 2) = 33

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 4 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 1 + 32

数字相加:

1 + 32 = 33

= 33

u_{1, 2} = \frac{- 1 \pm 33}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- 1 + 33}{2 \cdot 4}, u_{2} = \frac{- 1 - 33}{2 \cdot 4}

u = \frac{- 1 + 33}{2 \cdot 4} : \frac{- 1 + 33}{8}

\frac{- 1 + 33}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 + 33}{8}

u = \frac{- 1 - 33}{2 \cdot 4} : \frac{- 1 - 33}{8}

\frac{- 1 - 33}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 - 33}{8}

二次方程组的解是:

u = \frac{- 1 + 33}{8}, u = \frac{- 1 - 33}{8}

u = cos (x)

代回

cos (x) = \frac{- 1 + 33}{8}, cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 + 33}{8}, cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 + 33}{8} : x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

cos (x) = \frac{- 1 + 33}{8}

使用反三角函数性质

cos (x) = \frac{- 1 + 33}{8}

cos (x) = \frac{- 1 + 33}{8}

的通解

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

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn

cos (x) = \frac{- 1 - 33}{8} : x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

cos (x) = \frac{- 1 - 33}{8}

使用反三角函数性质

cos (x) = \frac{- 1 - 33}{8}

cos (x) = \frac{- 1 - 33}{8}

的通解

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

x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

合并所有解

x = arccos (\frac{- 1 + 33}{8}) + 2 πn, x = 2 π - arccos (\frac{- 1 + 33}{8}) + 2 πn, x = arccos (\frac{- 1 - 33}{8}) + 2 πn, x = - arccos (\frac{- 1 - 33}{8}) + 2 πn

以小数形式表示解

x = 0.93592 \dots + 2 πn, x = 2 π - 0.93592 \dots + 2 πn, x = 2.57376 \dots + 2 πn, x = - 2.57376 \dots + 2 πn

作图

流行的例子

tan(x)=-1+sqrt(2)

tan (x) = - 1 + 2

cot (θ) = - \frac{3}{4}

solvefor y,7e^x-sin(y)=x

so l v e f or y, 7 e^{x} - sin (y) = x

sin (B) = 0.4848

sin (x) = \frac{10}{12}