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受欢迎的三角函数 >

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

解答

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

解答

x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

+1

度数

x = 72.37560 \dots^{\circ} + 36 0^{\circ} n, x = 287.62439 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

u^{2} + 3 u = 1

u^{2} + 3 u = 1 : u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

u^{2} + 3 u = 1

将

1

para o lado esquerdo

u^{2} + 3 u = 1

两边减去

1

u^{2} + 3 u - 1 = 1 - 1

化简

u^{2} + 3 u - 1 = 0

u^{2} + 3 u - 1 = 0

使用求根公式求解

u^{2} + 3 u - 1 = 0

二次方程求根公式:

若

a = 1, b = 3, c = - 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 1) = 13

3^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 3^{2} + 4

3^{2} = 9

= 9 + 4

数字相加:

9 + 4 = 13

= 13

u_{1, 2} = \frac{- 3 \pm 13}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 3 + 13}{2 \cdot 1}, u_{2} = \frac{- 3 - 13}{2 \cdot 1}

u = \frac{- 3 + 13}{2 \cdot 1} : \frac{- 3 + 13}{2}

\frac{- 3 + 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 3 + 13}{2}

u = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

\frac{- 3 - 13}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 3 - 13}{2}

二次方程组的解是:

u = \frac{- 3 + 13}{2}, u = \frac{- 3 - 13}{2}

u = cos (x)

代回

cos (x) = \frac{- 3 + 13}{2}, cos (x) = \frac{- 3 - 13}{2}

cos (x) = \frac{- 3 + 13}{2}, cos (x) = \frac{- 3 - 13}{2}

cos (x) = \frac{- 3 + 13}{2} : x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

cos (x) = \frac{- 3 + 13}{2}

使用反三角函数性质

cos (x) = \frac{- 3 + 13}{2}

cos (x) = \frac{- 3 + 13}{2}

的通解

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

x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

cos (x) = \frac{- 3 - 13}{2} :

无解

cos (x) = \frac{- 3 - 13}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (\frac{- 3 + 13}{2}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 13}{2}) + 2 πn

以小数形式表示解

x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

作图

流行的例子

cot^2(x)=(tan(x))/2

cot^{2} (x) = \frac{tan ( x )}{2}

tan(θ)=(3.2)/(4.1)

tan (θ) = \frac{3.2}{4.1}

5sec(x)tan(x)=0

5 sec (x) tan (x) = 0

cos(5x)-cos(x)=2sin(2x)

cos (5 x) - cos (x) = 2 sin (2 x)

6arccos(4x)=5pi

6 arccos (4 x) = 5 π