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受欢迎的三角函数 >

1/(1-cos(x))+1/(1+cos(x))=2sec^2(x)

解答

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

解答

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2 sec^{2} (x)

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

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

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

= 2 - 2 sec^{2} (x) + 2 (\frac{1}{sec ( x )})^{2} sec^{2} (x)

化简

2 - 2 sec^{2} (x) + 2 (\frac{1}{sec ( x )})^{2} sec^{2} (x) : - 2 sec^{2} (x) + 4

2 - 2 sec^{2} (x) + 2 (\frac{1}{sec ( x )})^{2} sec^{2} (x)

2 (\frac{1}{sec ( x )})^{2} sec^{2} (x) = 2

2 (\frac{1}{sec ( x )})^{2} sec^{2} (x)

(\frac{1}{sec ( x )})^{2} = \frac{1}{sec ^{2} ( x )}

(\frac{1}{sec ( x )})^{2}

使用指数法则:

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

= \frac{1 ^{2}}{sec ^{2} ( x )}

使用法则

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( x )}

= 2 \cdot \frac{1}{sec ^{2} ( x )} sec^{2} (x)

分式相乘:

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

= \frac{1 \cdot 2 sec ^{2} ( x )}{sec ^{2} ( x )}

约分:

sec^{2} (x)

= 1 \cdot 2

数字相乘:

1 \cdot 2 = 2

= 2

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

对同类项分组

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

数字相加:

2 + 2 = 4

= - 2 sec^{2} (x) + 4

= - 2 sec^{2} (x) + 4

4 - 2 sec^{2} (x) = 0

用替代法求解

4 - 2 sec^{2} (x) = 0

令

: sec (x) = u

4 - 2 u^{2} = 0

4 - 2 u^{2} = 0 : u = 2, u = - 2

4 - 2 u^{2} = 0

将

4

到右边

4 - 2 u^{2} = 0

两边减去

4

4 - 2 u^{2} - 4 = 0 - 4

化简

- 2 u^{2} = - 4

- 2 u^{2} = - 4

两边除以

- 2

- 2 u^{2} = - 4

两边除以

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 4}{- 2}

化简

\frac{- 2 u ^{2}}{- 2} = \frac{- 4}{- 2}

化简

\frac{- 2 u ^{2}}{- 2} : u^{2}

\frac{- 2 u ^{2}}{- 2}

使用分式法则:

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

= \frac{2 u ^{2}}{2}

数字相除:

\frac{2}{2} = 1

= u^{2}

化简

\frac{- 4}{- 2} : 2

\frac{- 4}{- 2}

使用分式法则:

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

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

u^{2} = 2

u^{2} = 2

u^{2} = 2

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 2, u = - 2

u = sec (x)

代回

sec (x) = 2, sec (x) = - 2

sec (x) = 2, sec (x) = - 2

sec (x) = 2 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sec (x) = 2

sec (x) = 2

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sec (x) = - 2 : x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

sec (x) = - 2

sec (x) = - 2

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

合并所有解

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn

作图

流行的例子

2cos^2(θ)-sin(θ)=2

2 cos^{2} (θ) - sin (θ) = 2

cos(a)=(57)/(sqrt(113)*\sqrt{78)}

cos (a) = \frac{57}{113 \cdot 78}

tan^2(x)+tan(x)-12=2

tan^{2} (x) + tan (x) - 12 = 2

sin (o) = \frac{3}{5}

a cos (x) = cos (x)