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受欢迎的三角函数 >

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

解答

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

解答

x = 2.23703 \dots + 2 πn, x = - 2.23703 \dots + 2 πn

+1

度数

x = 128.17270 \dots^{\circ} + 36 0^{\circ} n, x = - 128.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

cos (2 x)

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

- (2 cos^{2} (x) - 1) + cos (x) + cos^{2} (x) : - cos^{2} (x) + cos (x) + 1

- (2 cos^{2} (x) - 1) + cos (x) + cos^{2} (x)

- (2 cos^{2} (x) - 1) : - 2 cos^{2} (x) + 1

- (2 cos^{2} (x) - 1)

打开括号

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

使用加减运算法则

- (- a) = a, - (a) = - a

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

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

化简

- 2 cos^{2} (x) + 1 + cos (x) + cos^{2} (x) : - cos^{2} (x) + cos (x) + 1

- 2 cos^{2} (x) + 1 + cos (x) + cos^{2} (x)

对同类项分组

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

同类项相加:

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

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

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

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

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

用替代法求解

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

令

: cos (x) = u

1 + u - u^{2} = 0

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

1 + u - u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

u = \frac{- 1 + 5}{2 ( - 1 )} : - \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{- 2}

使用分式法则:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 ( - 1 )} : \frac{1 + 5}{2}

\frac{- 1 - 5}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 5}{- 2}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

- 1 - 5 = - (1 + 5)

= \frac{1 + 5}{2}

二次方程组的解是:

u = - \frac{- 1 + 5}{2}, u = \frac{1 + 5}{2}

u = cos (x)

代回

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}, cos (x) = \frac{1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2} : x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = - \frac{- 1 + 5}{2}

使用反三角函数性质

cos (x) = - \frac{- 1 + 5}{2}

cos (x) = - \frac{- 1 + 5}{2}

的通解

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

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

cos (x) = \frac{1 + 5}{2} :

无解

cos (x) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = arccos (- \frac{- 1 + 5}{2}) + 2 πn, x = - arccos (- \frac{- 1 + 5}{2}) + 2 πn

以小数形式表示解

x = 2.23703 \dots + 2 πn, x = - 2.23703 \dots + 2 πn

作图

流行的例子

cos(x)= 3/(sqrt(13))

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

3tan^3(x)-9tan(x)=0

3 tan^{3} (x) - 9 tan (x) = 0

cot(θ)=-1,cos(θ),90<θ<180

cot (θ) = - 1, cos (θ), 9 0^{\circ} < θ < 18 0^{\circ}

2sin(3θ)+1=0,0<= θ<= 2pi

2 sin (3 θ) + 1 = 0, 0 \leq θ \leq 2 π

arctan(x)=-pi/6

arctan (x) = - \frac{π}{6}