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受欢迎的三角函数 >

4cos^2(x)+4cos(x)=1,0<= x<360

解答

4 cos^{2} (x) + 4 cos (x) = 1, 0^{\circ} \leq x < 36 0^{\circ}

解答

x = 1.36217 \dots, x = 36 0^{\circ} - 1.36217 \dots

+1

弧度

x = 1.36217 \dots, x = 2 π - 1.36217 \dots

求解步骤

4 cos^{2} (x) + 4 cos (x) = 1, 0^{\circ} \leq x < 36 0^{\circ}

用替代法求解

4 cos^{2} (x) + 4 cos (x) = 1

令

: cos (x) = u

4 u^{2} + 4 u = 1

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

4 u^{2} + 4 u = 1

将

1

para o lado esquerdo

4 u^{2} + 4 u = 1

两边减去

1

4 u^{2} + 4 u - 1 = 1 - 1

化简

4 u^{2} + 4 u - 1 = 0

4 u^{2} + 4 u - 1 = 0

使用求根公式求解

4 u^{2} + 4 u - 1 = 0

二次方程求根公式:

若

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

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

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

4^{2} - 4 \cdot 4 (- 1) = 42

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} + 16

4^{2} = 16

= 16 + 16

数字相加:

16 + 16 = 32

= 32

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{4}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

整理后得

= 42

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 + 4 2}{8}

分解

- 4 + 42 : 4 (- 1 + 2)

- 4 + 42

改写为

= - 4 \cdot 1 + 42

因式分解出通项

4

= 4 (- 1 + 2)

= \frac{4 ( - 1 + 2 )}{8}

约分:

4

= \frac{- 1 + 2}{2}

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

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 - 4 2}{8}

分解

- 4 - 42 : - 4 (1 + 2)

- 4 - 42

改写为

= - 4 \cdot 1 - 42

因式分解出通项

4

= - 4 (1 + 2)

= - \frac{4 ( 1 + 2 )}{8}

约分:

4

= - \frac{1 + 2}{2}

二次方程组的解是:

u = \frac{- 1 + 2}{2}, u = - \frac{1 + 2}{2}

u = cos (x)

代回

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

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

cos (x) = \frac{- 1 + 2}{2}, 0 \leq x < 36 0^{\circ} : x = arccos (\frac{2 - 1}{2}), x = 36 0^{\circ} - arccos (\frac{2 - 1}{2})

cos (x) = \frac{- 1 + 2}{2}, 0 \leq x < 36 0^{\circ}

使用反三角函数性质

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

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

的通解

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

x = arccos (\frac{- 1 + 2}{2}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{- 1 + 2}{2}) + 36 0^{\circ} n

x = arccos (\frac{- 1 + 2}{2}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{- 1 + 2}{2}) + 36 0^{\circ} n

在

0 \leq x < 36 0^{\circ}

范围内的解

x = arccos (\frac{2 - 1}{2}), x = 36 0^{\circ} - arccos (\frac{2 - 1}{2})

cos (x) = - \frac{1 + 2}{2}, 0 \leq x < 36 0^{\circ} :

无解

cos (x) = - \frac{1 + 2}{2}, 0 \leq x < 36 0^{\circ}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (\frac{2 - 1}{2}), x = 36 0^{\circ} - arccos (\frac{2 - 1}{2})

以小数形式表示解

x = 1.36217 \dots, x = 36 0^{\circ} - 1.36217 \dots

作图

流行的例子

8 cos^{2} (x) = 4

4 cos (3 x) = 0

sin^2(x)=sin^2(x/2)

sin^{2} (x) = sin^{2} (\frac{x}{2})

cos (x) = \frac{6}{8}

3 sin (x) = 2