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受欢迎的三角函数 >

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

解答

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

解答

x = 0.74946 \dots + 2 πn, x = 2 π - 0.74946 \dots + 2 πn

+1

度数

x = 42.94140 \dots^{\circ} + 36 0^{\circ} n, x = 317.05859 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

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

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

2^{2} - 4 \cdot 1 \cdot (- 2) = 23

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

数字相加:

4 + 8 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

分解

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

改写为

= - 2 \cdot 1 + 23

因式分解出通项

2

= 2 (- 1 + 3)

= \frac{2 ( - 1 + 3 )}{2}

数字相除:

\frac{2}{2} = 1

= - 1 + 3

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

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

数字相乘:

2 \cdot 1 = 2

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

分解

- 2 - 23 : - 2 (1 + 3)

- 2 - 23

改写为

= - 2 \cdot 1 - 23

因式分解出通项

2

= - 2 (1 + 3)

= - \frac{2 ( 1 + 3 )}{2}

数字相除:

\frac{2}{2} = 1

= - (1 + 3)

取负

- (1 + 3) = - 1 - 3

= - 1 - 3

二次方程组的解是:

u = - 1 + 3, u = - 1 - 3

u = cos (x)

代回

cos (x) = - 1 + 3, cos (x) = - 1 - 3

cos (x) = - 1 + 3, cos (x) = - 1 - 3

cos (x) = - 1 + 3 : x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

cos (x) = - 1 + 3

使用反三角函数性质

cos (x) = - 1 + 3

cos (x) = - 1 + 3

的通解

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

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

cos (x) = - 1 - 3 :

无解

cos (x) = - 1 - 3

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- 1 + 3) + 2 πn, x = 2 π - arccos (- 1 + 3) + 2 πn

以小数形式表示解

x = 0.74946 \dots + 2 πn, x = 2 π - 0.74946 \dots + 2 πn

作图

流行的例子

- sin (4 x) \cdot 4 = 0

cos (θ) = \frac{5}{6}

cos(2x)=cos^2(x)-1/2

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

cos(x)=1,0<= x<= 2pi

cos (x) = 1, 0 \leq x \leq 2 π

solvefor x,cot(-x)cos(-x)+sin(-x)=(-1)/f

so l v e f or x, cot (- x) cos (- x) + sin (- x) = \frac{- 1}{f}