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受欢迎的三角函数 >

tan^2(2x)-1=sec(2x)

解答

tan^{2} (2 x) - 1 = sec (2 x)

解答

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn, x = \frac{π}{2} + πn

+1

度数

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

求解步骤

tan^{2} (2 x) - 1 = sec (2 x)

两边减去

sec (2 x)

tan^{2} (2 x) - 1 - sec (2 x) = 0

使用三角恒等式改写

- 1 - sec (2 x) + tan^{2} (2 x)

使用毕达哥拉斯恒等式:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

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

化简

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

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

对同类项分组

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

数字相减:

- 1 - 1 = - 2

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

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

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

用替代法求解

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

令

: sec (2 x) = u

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

- 2 - u + u^{2} = 0 : u = 2, u = - 1

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

改写成标准形式

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

u^{2} - u - 2 = 0

使用求根公式求解

u^{2} - u - 2 = 0

二次方程求根公式:

若

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

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

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

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

使用法则

- (- a) = a

= (- 1)^{2} + 4 \cdot 1 \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 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数字相乘:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 1}, u_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 1}

u = \frac{- ( - 1 ) + 3}{2 \cdot 1} : 2

\frac{- ( - 1 ) + 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 3}{2 \cdot 1}

数字相加:

1 + 3 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

u = \frac{- ( - 1 ) - 3}{2 \cdot 1} : - 1

\frac{- ( - 1 ) - 3}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 - 3}{2 \cdot 1}

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 2}{2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = 2, u = - 1

u = sec (2 x)

代回

sec (2 x) = 2, sec (2 x) = - 1

sec (2 x) = 2, sec (2 x) = - 1

sec (2 x) = 2 : x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

sec (2 x) = 2

sec (2 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}

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

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

解

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

2 x = \frac{π}{3} + 2 πn

两边除以

2

2 x = \frac{π}{3} + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 x}{2} = \frac{\frac{π}{3}}{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{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

解

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{3} + 2 πn

两边除以

2

2 x = \frac{5 π}{3} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{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{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

使用分式法则:

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

= \frac{5 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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

sec (2 x) = - 1

sec (2 x) = - 1

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}

2 x = π + 2 πn

2 x = π + 2 πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

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

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

合并所有解

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn, x = \frac{π}{2} + πn

作图

流行的例子

- 1.92 = cos (x)

5cot^2(x)+8cot(x)+3=0

5 cot^{2} (x) + 8 cot (x) + 3 = 0

sin (x) = \frac{15}{31}

4 cot (θ) - 1 = 0

solvefor x,4cos^2(x)=5-4sin(x)

so l v e f or x, 4 cos^{2} (x) = 5 - 4 sin (x)