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受欢迎的三角函数 >

tan(2x)+sin(2x)+cos(2x)=sec(2x)

解答

tan (2 x) + sin (2 x) + cos (2 x) = sec (2 x)

解答

x = \frac{π + 2 πn}{2}, x = πn, x = \frac{4 πn + π}{4}

+1

度数

x = 9 0^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

tan (2 x) + sin (2 x) + cos (2 x) = sec (2 x)

两边减去

sec (2 x)

tan (2 x) + sin (2 x) + cos (2 x) - sec (2 x) = 0

令

: u = 2 x

tan (u) + sin (u) + cos (u) - sec (u) = 0

用 sin, cos 表示

cos (u) - sec (u) + sin (u) + tan (u)

使用基本三角恒等式:

sec (x) = \frac{1}{cos ( x )}

= cos (u) - \frac{1}{cos ( u )} + sin (u) + tan (u)

使用基本三角恒等式:

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

= cos (u) - \frac{1}{cos ( u )} + sin (u) + \frac{sin ( u )}{cos ( u )}

化简

cos (u) - \frac{1}{cos ( u )} + sin (u) + \frac{sin ( u )}{cos ( u )} : \frac{cos ^{2} ( u ) - 1 + sin ( u ) + sin ( u ) cos ( u )}{cos ( u )}

cos (u) - \frac{1}{cos ( u )} + sin (u) + \frac{sin ( u )}{cos ( u )}

合并分式

- \frac{1}{cos ( u )} + \frac{sin ( u )}{cos ( u )} : \frac{- 1 + sin ( u )}{cos ( u )}

使用法则

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

= \frac{- 1 + sin ( u )}{cos ( u )}

= cos (u) + \frac{sin ( u ) - 1}{cos ( u )} + sin (u)

将项转换为分式:

cos (u) = \frac{cos ( u ) cos ( u )}{cos ( u )}, sin (u) = \frac{sin ( u ) cos ( u )}{cos ( u )}

= \frac{cos ( u ) cos ( u )}{cos ( u )} + \frac{- 1 + sin ( u )}{cos ( u )} + \frac{sin ( u ) cos ( u )}{cos ( u )}

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 ( u ) cos ( u ) - 1 + sin ( u ) + sin ( u ) cos ( u )}{cos ( u )}

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

cos (u) cos (u) - 1 + sin (u) + sin (u) cos (u)

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

cos (u) cos (u)

使用指数法则:

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

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

= cos^{1 + 1} (u)

数字相加:

1 + 1 = 2

= cos^{2} (u)

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

= \frac{cos ^{2} ( u ) - 1 + sin ( u ) + sin ( u ) cos ( u )}{cos ( u )}

= \frac{cos ^{2} ( u ) - 1 + sin ( u ) + sin ( u ) cos ( u )}{cos ( u )}

\frac{- 1 + cos ^{2} ( u ) + sin ( u ) + cos ( u ) sin ( u )}{cos ( u )} = 0

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

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

使用三角恒等式改写

- 1 + cos^{2} (u) + sin (u) + cos (u) sin (u)

使用毕达哥拉斯恒等式:

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

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

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

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

sin^{2} (u)

使用毕达哥拉斯恒等式:

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

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

= 1 - cos^{2} (u)

分解

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

1 - cos^{2} (u)

使用平方差公式:

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

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

= (1 + cos (u)) (1 - cos (u))

= (1 + cos (u)) (1 - cos (u))

= sin (u) - (1 + cos (u)) (1 - cos (u)) + cos (u) sin (u)

sin (u) - (1 + cos (u)) (1 - cos (u)) + cos (u) sin (u) = 0

分解

sin (u) - (1 + cos (u)) (1 - cos (u)) + cos (u) sin (u) : (1 + cos (u)) (sin (u) + cos (u) - 1)

sin (u) - (1 + cos (u)) (1 - cos (u)) + cos (u) sin (u)

因式分解出通项

sin (u)

= sin (u) (1 + cos (u)) - (1 + cos (u)) (1 - cos (u))

因式分解出通项

(1 + cos (u))

= (1 + cos (u)) (sin (u) - (1 - cos (u)))

分解

sin (u) - (- cos (u) + 1) : sin (u) + cos (u) - 1

sin (u) - (1 - cos (u))

- (1 - cos (u)) = cos (u) - 1

- (1 - cos (u))

分解

1 - cos (u) : - (cos (u) - 1)

1 - cos (u)

因式分解出通项

- 1

= - (cos (u) - 1)

= (cos (u) - 1)

整理后得

= cos (u) - 1

= sin (u) + cos (u) - 1

= (cos (u) + 1) (sin (u) + cos (u) - 1)

(1 + cos (u)) (sin (u) + cos (u) - 1) = 0

分别求解每个部分

1 + cos (u) = 0 or sin (u) + cos (u) - 1 = 0

1 + cos (u) = 0 : u = π + 2 πn

1 + cos (u) = 0

将

1

到右边

1 + cos (u) = 0

两边减去

1

1 + cos (u) - 1 = 0 - 1

化简

cos (u) = - 1

cos (u) = - 1

cos (u) = - 1

的通解

cos (x)

周期表（周期为

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

u = π + 2 πn

u = π + 2 πn

sin (u) + cos (u) - 1 = 0 : u = 2 πn, u = 2 πn + \frac{π}{2}

sin (u) + cos (u) - 1 = 0

使用三角恒等式改写

sin (u) + cos (u) - 1

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

sin (u) + cos (u)

改写为

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

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 (u) + sin (\frac{π}{4}) cos (u))

使用角和恒等式:

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

= 2 sin (u + \frac{π}{4})

= - 1 + 2 sin (u + \frac{π}{4})

- 1 + 2 sin (u + \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (u + \frac{π}{4}) = 0

两边加上

1

- 1 + 2 sin (u + \frac{π}{4}) + 1 = 0 + 1

化简

2 sin (u + \frac{π}{4}) = 1

2 sin (u + \frac{π}{4}) = 1

两边除以

2

2 sin (u + \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( u + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( u + \frac{π}{4} )}{2} = \frac{1}{2}

化简

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

\frac{2 sin ( u + \frac{π}{4} )}{2}

约分:

2

= sin (u + \frac{π}{4})

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (u + \frac{π}{4}) = \frac{2}{2}

sin (u + \frac{π}{4}) = \frac{2}{2}

sin (u + \frac{π}{4}) = \frac{2}{2}

sin (u + \frac{π}{4}) = \frac{2}{2}

的通解

sin (x)

周期表（周期为

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

u + \frac{π}{4} = \frac{π}{4} + 2 πn, u + \frac{π}{4} = \frac{3 π}{4} + 2 πn

u + \frac{π}{4} = \frac{π}{4} + 2 πn, u + \frac{π}{4} = \frac{3 π}{4} + 2 πn

解

u + \frac{π}{4} = \frac{π}{4} + 2 πn : u = 2 πn

u + \frac{π}{4} = \frac{π}{4} + 2 πn

两边减去

\frac{π}{4}

u + \frac{π}{4} - \frac{π}{4} = \frac{π}{4} + 2 πn - \frac{π}{4}

化简

u = 2 πn

解

u + \frac{π}{4} = \frac{3 π}{4} + 2 πn : u = 2 πn + \frac{π}{2}

u + \frac{π}{4} = \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

u + \frac{π}{4} = \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{4}

u + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

u + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

u + \frac{π}{4} - \frac{π}{4} : u

u + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= u

化简

\frac{3 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{2}

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

u = 2 πn + \frac{π}{2}

u = 2 πn + \frac{π}{2}

u = 2 πn + \frac{π}{2}

u = 2 πn, u = 2 πn + \frac{π}{2}

合并所有解

u = π + 2 πn, u = 2 πn, u = 2 πn + \frac{π}{2}

u = 2 x

代回

2 x = π + 2 πn : x = \frac{π + 2 πn}{2}

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{π}{2} + \frac{2 πn}{2} : \frac{π + 2 πn}{2}

\frac{π}{2} + \frac{2 πn}{2}

使用法则

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

= \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

x = \frac{π + 2 πn}{2}

2 x = 2 πn : x = πn

2 x = 2 πn

两边除以

2

2 x = 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2}

化简

x = πn

x = πn

2 x = 2 πn + \frac{π}{2} : x = \frac{4 πn + π}{4}

2 x = 2 πn + \frac{π}{2}

两边除以

2

2 x = 2 πn + \frac{π}{2}

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

化简

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 πn}{2} + \frac{\frac{π}{2}}{2} : \frac{4 πn + π}{4}

\frac{2 πn}{2} + \frac{\frac{π}{2}}{2}

使用法则

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

= \frac{2 πn + \frac{π}{2}}{2}

化简

2 πn + \frac{π}{2} : \frac{4 πn + π}{2}

2 πn + \frac{π}{2}

将项转换为分式:

2 πn = \frac{2 πn 2}{2}

= \frac{2 πn \cdot 2}{2} + \frac{π}{2}

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 πn \cdot 2 + π}{2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 πn + π}{2}

= \frac{\frac{4 πn + π}{2}}{2}

使用分式法则:

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

= \frac{4 πn + π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{4 πn + π}{4}

x = \frac{4 πn + π}{4}

x = \frac{4 πn + π}{4}

x = \frac{4 πn + π}{4}

x = \frac{π + 2 πn}{2}, x = πn, x = \frac{4 πn + π}{4}

因为方程对以下值无定义:

\frac{4 πn + π}{4}

x = \frac{π + 2 πn}{2}, x = πn, x = \frac{4 πn + π}{4}

作图

流行的例子

1-cos(4x)=sin(2x)

1 - cos (4 x) = sin (2 x)

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

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

sec^2(x)-3tan(x)=5

sec^{2} (x) - 3 tan (x) = 5

sin (x) = 0, 4

sin (x) = 0, 2