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受欢迎的三角函数 >

2cos(x)=tan(x)

解答

2 cos (x) = tan (x)

解答

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

+1

度数

x = 51.33171 \dots^{\circ} + 36 0^{\circ} n, x = 128.66828 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (x) = tan (x)

两边减去

tan (x)

2 cos (x) - tan (x) = 0

用 sin, cos 表示

2 cos (x) - \frac{sin ( x )}{cos ( x )} = 0

化简

2 cos (x) - \frac{sin ( x )}{cos ( x )} : \frac{2 cos ^{2} ( x ) - sin ( x )}{cos ( x )}

2 cos (x) - \frac{sin ( x )}{cos ( x )}

将项转换为分式:

2 cos (x) = \frac{2 cos ( x ) cos ( x )}{cos ( x )}

= \frac{2 cos ( x ) cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 cos ( x ) cos ( x ) - sin ( x )}{cos ( x )}

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

2 cos (x) cos (x) - sin (x)

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

2 cos (x) cos (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= 2 cos^{2} (x)

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

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

\frac{2 cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

两边加上

sin (x)

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

两边进行平方

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

两边减去

sin^{2} (x)

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

分解

4 cos^{4} (x) - sin^{2} (x) : (2 cos^{2} (x) + sin (x)) (2 cos^{2} (x) - sin (x))

4 cos^{4} (x) - sin^{2} (x)

将

4 cos^{4} (x) - sin^{2} (x)

改写为

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

4 cos^{4} (x) - sin^{2} (x)

将

4

改写为

2^{2}

= 2^{2} cos^{4} (x) - sin^{2} (x)

使用指数法则:

a^{b c} = (a^{b})^{c}

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

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

使用指数法则:

a^{m} b^{m} = (ab)^{m}

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

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

2 cos^{2} (x) + sin (x) = 0 : x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

sin (x) + (1 - sin^{2} (x)) \cdot 2 = 0

用替代法求解

sin (x) + (1 - sin^{2} (x)) \cdot 2 = 0

令

: sin (x) = u

u + (1 - u^{2}) \cdot 2 = 0

u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{- 1 + 17}{4}, u = \frac{1 + 17}{4}

u + (1 - u^{2}) \cdot 2 = 0

展开

u + (1 - u^{2}) \cdot 2 : u + 2 - 2 u^{2}

u + (1 - u^{2}) \cdot 2

= u + 2 (1 - u^{2})

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

= u + 2 - 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 4 (- 2) \cdot 2 = 17

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 2

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 1 + 17}{2 ( - 2 )}, u_{2} = \frac{- 1 - 17}{2 ( - 2 )}

u = \frac{- 1 + 17}{2 ( - 2 )} : - \frac{- 1 + 17}{4}

\frac{- 1 + 17}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 + 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 + 17}{- 4}

使用分式法则:

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

= - \frac{- 1 + 17}{4}

u = \frac{- 1 - 17}{2 ( - 2 )} : \frac{1 + 17}{4}

\frac{- 1 - 17}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 1 - 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 1 - 17}{- 4}

使用分式法则:

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

- 1 - 17 = - (1 + 17)

= \frac{1 + 17}{4}

二次方程组的解是:

u = - \frac{- 1 + 17}{4}, u = \frac{1 + 17}{4}

u = sin (x)

代回

sin (x) = - \frac{- 1 + 17}{4}, sin (x) = \frac{1 + 17}{4}

sin (x) = - \frac{- 1 + 17}{4}, sin (x) = \frac{1 + 17}{4}

sin (x) = - \frac{- 1 + 17}{4} : x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = - \frac{- 1 + 17}{4}

使用反三角函数性质

sin (x) = - \frac{- 1 + 17}{4}

sin (x) = - \frac{- 1 + 17}{4}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{1 + 17}{4} :

无解

sin (x) = \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

2 cos^{2} (x) - sin (x) = 0 : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

- sin (x) + (1 - sin^{2} (x)) \cdot 2 = 0

用替代法求解

- sin (x) + (1 - sin^{2} (x)) \cdot 2 = 0

令

: sin (x) = u

- u + (1 - u^{2}) \cdot 2 = 0

- u + (1 - u^{2}) \cdot 2 = 0 : u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

- u + (1 - u^{2}) \cdot 2 = 0

展开

- u + (1 - u^{2}) \cdot 2 : - u + 2 - 2 u^{2}

- u + (1 - u^{2}) \cdot 2

= - u + 2 (1 - u^{2})

乘开

2 (1 - u^{2}) : 2 - 2 u^{2}

2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = u^{2}

= 2 \cdot 1 - 2 u^{2}

数字相乘:

2 \cdot 1 = 2

= 2 - 2 u^{2}

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 1)^{2} - 4 (- 2) \cdot 2 = 17

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

数字相加:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 17}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 17}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 17}{2 ( - 2 )} : - \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 ( - 2 )}

去除括号:

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

= \frac{1 + 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 + 17}{- 4}

使用分式法则:

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

= - \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 ( - 2 )} : \frac{17 - 1}{4}

\frac{- ( - 1 ) - 17}{2 ( - 2 )}

去除括号:

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

= \frac{1 - 17}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 - 17}{- 4}

使用分式法则:

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

1 - 17 = - (17 - 1)

= \frac{17 - 1}{4}

二次方程组的解是:

u = - \frac{1 + 17}{4}, u = \frac{17 - 1}{4}

u = sin (x)

代回

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4}, sin (x) = \frac{17 - 1}{4}

sin (x) = - \frac{1 + 17}{4} :

无解

sin (x) = - \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

无解

sin (x) = \frac{17 - 1}{4} : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

sin (x) = \frac{17 - 1}{4}

使用反三角函数性质

sin (x) = \frac{17 - 1}{4}

sin (x) = \frac{17 - 1}{4}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

合并所有解

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

合并所有解

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

将解代入原方程进行验证

将它们代入

2 cos (x) = tan (x)

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{- 1 + 17}{4}) + 2 πn

的解:

假

arcsin (- \frac{- 1 + 17}{4}) + 2 πn

代入

n = 1

arcsin (- \frac{- 1 + 17}{4}) + 2 π 1

对于

2 cos (x) = tan (x)

代入

x = arcsin (- \frac{- 1 + 17}{4}) + 2 π 1

2 cos (arcsin (- \frac{- 1 + 17}{4}) + 2 π 1) = tan (arcsin (- \frac{- 1 + 17}{4}) + 2 π 1)

整理后得

1.24962 \dots = - 1.24962 \dots

\Rightarrow 假

检验

π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

的解:

假

π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

代入

n = 1

π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1

对于

2 cos (x) = tan (x)

代入

x = π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1

2 cos (π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1) = tan (π + arcsin (\frac{- 1 + 17}{4}) + 2 π 1)

整理后得

- 1.24962 \dots = 1.24962 \dots

\Rightarrow 假

检验

arcsin (\frac{17 - 1}{4}) + 2 πn

的解:

真

arcsin (\frac{17 - 1}{4}) + 2 πn

代入

n = 1

arcsin (\frac{17 - 1}{4}) + 2 π 1

对于

2 cos (x) = tan (x)

代入

x = arcsin (\frac{17 - 1}{4}) + 2 π 1

2 cos (arcsin (\frac{17 - 1}{4}) + 2 π 1) = tan (arcsin (\frac{17 - 1}{4}) + 2 π 1)

整理后得

1.24962 \dots = 1.24962 \dots

\Rightarrow 真

检验

π - arcsin (\frac{17 - 1}{4}) + 2 πn

的解:

真

π - arcsin (\frac{17 - 1}{4}) + 2 πn

代入

n = 1

π - arcsin (\frac{17 - 1}{4}) + 2 π 1

对于

2 cos (x) = tan (x)

代入

x = π - arcsin (\frac{17 - 1}{4}) + 2 π 1

2 cos (π - arcsin (\frac{17 - 1}{4}) + 2 π 1) = tan (π - arcsin (\frac{17 - 1}{4}) + 2 π 1)

整理后得

- 1.24962 \dots = - 1.24962 \dots

\Rightarrow 真

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

以小数形式表示解

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

作图

流行的例子

sin^3(a)=3sin(a)-4sin^3(a)

sin^{3} (a) = 3 sin (a) - 4 sin^{3} (a)

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

solvefor x,arcsin(y)-arccos(x)=1

so l v e f or x, arcsin (y) - arccos (x) = 1

-6=-2+8sin(θ+pi/4)

- 6 = - 2 + 8 sin (θ + \frac{π}{4})

0=12cos(pi/3 x)+8

0 = 12 cos (\frac{π}{3} x) + 8