zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

tan (2 x) = cos (2 x), 0 \leq x \leq π

解答

x = \frac{0.66623 \dots}{2}, x = \frac{π - 0.66623 \dots}{2}

+1

度数

x = 19.08635 \dots^{\circ}, x = 70.91364 \dots^{\circ}

求解步骤

tan (2 x) = cos (2 x), 0 \leq x \leq π

两边减去

cos (2 x)

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

用 sin, cos 表示

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

化简

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

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

将项转换为分式:

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

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

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

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

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

cos (2 x) cos (2 x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= cos^{2} (2 x)

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

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

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

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

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

两边加上

cos^{2} (2 x)

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

两边进行平方

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

两边减去

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

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

分解

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

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

使用指数法则:

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

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

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

使用平方差公式:

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

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

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

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

分别求解每个部分

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

sin (2 x) + cos^{2} (2 x) = 0, 0 \leq x \leq π : x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

sin (2 x) + cos^{2} (2 x) = 0, 0 \leq x \leq π

使用三角恒等式改写

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

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

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

用替代法求解

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

令

: sin (2 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 = sin (2 x)

代回

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

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

sin (2 x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq π : x = \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

sin (2 x) = - \frac{- 1 + 5}{2}, 0 \leq x \leq π

使用反三角函数性质

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

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

的通解

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

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

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

解

2 x = arcsin (- \frac{- 1 + 5}{2}) + 2 πn : x = - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

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

化简

arcsin (- \frac{- 1 + 5}{2}) + 2 πn : - arcsin (\frac{5 - 1}{2}) + 2 πn

arcsin (- \frac{- 1 + 5}{2}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

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

= - arcsin (\frac{5 - 1}{2}) + 2 πn

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

两边除以

2

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

两边除以

2

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

化简

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

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

解

2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn : x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

两边除以

2

2 x = π + arcsin (\frac{- 1 + 5}{2}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = \frac{π}{2} + \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

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

在

0 \leq x \leq π

范围内的解

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

sin (2 x) = \frac{1 + 5}{2}, 0 \leq x \leq π :

无解

sin (2 x) = \frac{1 + 5}{2}, 0 \leq x \leq π

- 1 \leq sin (x) \leq 1

无解

合并所有解

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

sin (2 x) - cos^{2} (2 x) = 0, 0 \leq x \leq π : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

sin (2 x) - cos^{2} (2 x) = 0, 0 \leq x \leq π

使用三角恒等式改写

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

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

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

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

打开括号

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

使用加减运算法则

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

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

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

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

用替代法求解

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

令

: sin (2 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 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 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 \cdot 1}

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

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

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

数字相乘:

2 \cdot 1 = 2

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

二次方程组的解是:

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

u = sin (2 x)

代回

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

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

sin (2 x) = \frac{- 1 + 5}{2}, 0 \leq x \leq π : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

sin (2 x) = \frac{- 1 + 5}{2}, 0 \leq x \leq π

使用反三角函数性质

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

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

的通解

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

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

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

解

2 x = arcsin (\frac{- 1 + 5}{2}) + 2 πn : x = \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

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

两边除以

2

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

两边除以

2

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

化简

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

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

解

2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{- 1 + 5}{2} )}{2} + πn

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

在

0 \leq x \leq π

范围内的解

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

sin (2 x) = \frac{- 1 - 5}{2}, 0 \leq x \leq π :

无解

sin (2 x) = \frac{- 1 - 5}{2}, 0 \leq x \leq π

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

合并所有解

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

将解代入原方程进行验证

将它们代入

tan (2 x) = cos (2 x)

检验解是否符合

去除与方程不符的解。

检验

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

的解:

假

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

代入

n = 1

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

对于

tan (2 x) = cos (2 x)

代入

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

tan 2 \cdot \frac{π + arcsin ( \frac{5 - 1}{2} )}{2} = cos 2 \cdot \frac{π + arcsin ( \frac{5 - 1}{2} )}{2}

整理后得

0.78615 \dots = - 0.78615 \dots

\Rightarrow 假

检验

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

的解:

假

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

代入

n = 1

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

对于

tan (2 x) = cos (2 x)

代入

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

tan 2 \cdot \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2} = cos 2 \cdot \frac{- arcsin ( \frac{5 - 1}{2} ) + 2 π}{2}

整理后得

- 0.78615 \dots = 0.78615 \dots

\Rightarrow 假

检验

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

的解:

真

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

代入

n = 1

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

对于

tan (2 x) = cos (2 x)

代入

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

tan 2 \cdot \frac{arcsin ( \frac{5 - 1}{2} )}{2} = cos 2 \cdot \frac{arcsin ( \frac{5 - 1}{2} )}{2}

整理后得

0.78615 \dots = 0.78615 \dots

\Rightarrow 真

检验

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

的解:

真

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

代入

n = 1

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

对于

tan (2 x) = cos (2 x)

代入

x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

tan 2 \cdot \frac{π - arcsin ( \frac{5 - 1}{2} )}{2} = cos 2 \cdot \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

整理后得

- 0.78615 \dots = - 0.78615 \dots

\Rightarrow 真

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2}, x = \frac{π - arcsin ( \frac{5 - 1}{2} )}{2}

以小数形式表示解

x = \frac{0.66623 \dots}{2}, x = \frac{π - 0.66623 \dots}{2}

作图

流行的例子

5sin(2x)=5cos(x),0<= x<= 2pi

5 sin (2 x) = 5 cos (x), 0 \leq x \leq 2 π

solvefor x,f=cos(x)cos(hy)

so l v e f or x, f = cos (x) cos (h y)

tan (x) = - \frac{1}{10}

tan(4x)*cot(x+60)=1

tan (4 x) \cdot cot (x + 60) = 1

sin^2(A)+cos^2(A)+sin(A)-2=0

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