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受欢迎的三角函数 >

solvefor x,f=(cos(2x))/(cos(x)-sin(x))

解答

求解

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

解答

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

求解步骤

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

交换两边

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

两边减去

f

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

化简

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

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

将项转换为分式:

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

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

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

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

cos (2 x) - f (cos (x) - sin (x)) = 0

使用三角恒等式改写

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

cos (2 x) = (cos (x) + sin (x)) (cos (x) - sin (x))

cos (2 x)

使用倍角公式:

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

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

分解

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

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

使用平方差公式:

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

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

= (cos (x) + sin (x)) (cos (x) - sin (x))

= (cos (x) + sin (x)) (cos (x) - sin (x))

= (cos (x) + sin (x)) (cos (x) - sin (x)) - f (cos (x) - sin (x))

(cos (x) + sin (x)) (cos (x) - sin (x)) - (cos (x) - sin (x)) f = 0

分解

(cos (x) + sin (x)) (cos (x) - sin (x)) - (cos (x) - sin (x)) f : (cos (x) - sin (x)) (cos (x) + sin (x) - f)

(cos (x) + sin (x)) (cos (x) - sin (x)) - (cos (x) - sin (x)) f

改写为

= (cos (x) - sin (x)) (cos (x) + sin (x)) - (cos (x) - sin (x)) f

因式分解出通项

(cos (x) - sin (x))

= (cos (x) - sin (x)) (cos (x) + sin (x) - f)

(cos (x) - sin (x)) (cos (x) + sin (x) - f) = 0

分别求解每个部分

cos (x) - sin (x) = 0 or cos (x) + sin (x) - f = 0

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

cos (x) - sin (x) = 0

使用三角恒等式改写

cos (x) - sin (x) = 0

在两边除以

cos (x), cos (x) \neq = 0

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

化简

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

使用基本三角恒等式:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (x) = 0

1 - tan (x) = 0

将

1

到右边

1 - tan (x) = 0

两边减去

1

1 - tan (x) - 1 = 0 - 1

化简

- tan (x) = - 1

- tan (x) = - 1

两边除以

- 1

- tan (x) = - 1

两边除以

- 1

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

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

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

cos (x) + sin (x) - f = 0

使用三角恒等式改写

cos (x) + sin (x) - f

sin (x) + cos (x) = 2 sin (x + \frac{π}{4})

sin (x) + cos (x)

改写为

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

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (x) + sin (\frac{π}{4}) cos (x))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (x + \frac{π}{4})

= - f + 2 sin (x + \frac{π}{4})

- f + 2 sin (x + \frac{π}{4}) = 0

将

f

到右边

- f + 2 sin (x + \frac{π}{4}) = 0

两边加上

f

- f + 2 sin (x + \frac{π}{4}) + f = 0 + f

化简

2 sin (x + \frac{π}{4}) = f

2 sin (x + \frac{π}{4}) = f

两边除以

2

2 sin (x + \frac{π}{4}) = f

两边除以

2

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{f}{2}

化简

\frac{2 sin ( x + \frac{π}{4} )}{2} = \frac{f}{2}

化简

\frac{2 sin ( x + \frac{π}{4} )}{2} : sin (x + \frac{π}{4})

\frac{2 sin ( x + \frac{π}{4} )}{2}

约分:

2

= sin (x + \frac{π}{4})

化简

\frac{f}{2} : \frac{2 f}{2}

\frac{f}{2}

乘以共轭根式

\frac{2}{2}

= \frac{f 2}{2 2}

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2 f}{2}

sin (x + \frac{π}{4}) = \frac{2 f}{2}

sin (x + \frac{π}{4}) = \frac{2 f}{2}

sin (x + \frac{π}{4}) = \frac{2 f}{2}

使用反三角函数性质

sin (x + \frac{π}{4}) = \frac{2 f}{2}

sin (x + \frac{π}{4}) = \frac{2 f}{2}

的通解

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

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

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

解

x + \frac{π}{4} = arcsin (\frac{2 f}{2}) + 2 πn : x = arcsin (\frac{f}{2}) + 2 πn - \frac{π}{4}

x + \frac{π}{4} = arcsin (\frac{2 f}{2}) + 2 πn

化简

arcsin (\frac{2 f}{2}) + 2 πn : arcsin (\frac{f}{2}) + 2 πn

arcsin (\frac{2 f}{2}) + 2 πn

\frac{2 f}{2} = \frac{f}{2}

\frac{2 f}{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} f}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{f}{2 ^{1 - \frac{1}{2}}}

数字相减:

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

= \frac{f}{2 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= \frac{f}{2}

= arcsin (\frac{f}{2}) + 2 πn

x + \frac{π}{4} = arcsin (\frac{f}{2}) + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = arcsin (\frac{f}{2}) + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = arcsin (\frac{f}{2}) + 2 πn - \frac{π}{4}

化简

x = arcsin (\frac{f}{2}) + 2 πn - \frac{π}{4}

x = arcsin (\frac{f}{2}) + 2 πn - \frac{π}{4}

解

x + \frac{π}{4} = π + arcsin (- \frac{2 f}{2}) + 2 πn : x = π + arcsin (- \frac{f}{2}) + 2 πn - \frac{π}{4}

x + \frac{π}{4} = π + arcsin (- \frac{2 f}{2}) + 2 πn

化简

π + arcsin (- \frac{2 f}{2}) + 2 πn : π + arcsin (- \frac{f}{2}) + 2 πn

π + arcsin (- \frac{2 f}{2}) + 2 πn

\frac{2 f}{2} = \frac{f}{2}

\frac{2 f}{2}

使用根式运算法则:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} f}{2}

使用指数法则:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

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

= \frac{f}{2 ^{1 - \frac{1}{2}}}

数字相减:

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

= \frac{f}{2 ^{\frac{1}{2}}}

使用根式运算法则:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= \frac{f}{2}

= π + arcsin (- \frac{f}{2}) + 2 πn

x + \frac{π}{4} = π + arcsin (- \frac{f}{2}) + 2 πn

将

\frac{π}{4}

到右边

x + \frac{π}{4} = π + arcsin (- \frac{f}{2}) + 2 πn

两边减去

\frac{π}{4}

x + \frac{π}{4} - \frac{π}{4} = π + arcsin (- \frac{f}{2}) + 2 πn - \frac{π}{4}

化简

x = π + arcsin (- \frac{f}{2}) + 2 πn - \frac{π}{4}

x = π + arcsin (- \frac{f}{2}) + 2 πn - \frac{π}{4}

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

合并所有解

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

因为方程对以下值无定义:

\frac{π}{4} + πn

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

作图

流行的例子

cos (2 x) + 5 = 6.25

sin (x) = 4 sin (8)

csc (x) = \frac{7}{3}

3-cos(x)=3-sin(x)

3 - cos (x) = 3 - sin (x)

cos(t)= 1/3 ,0<t< 1/2 pi

cos (t) = \frac{1}{3}, 0 < t < \frac{1}{2} π