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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: sec (x) = u

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

\frac{2 u ^{2} - 1}{u ^{2}} = u^{2} : u = 1, u = - 1

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

在两边乘以

u^{2}

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

在两边乘以

u^{2}

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

化简

u^{2} u^{2} : u^{4}

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

使用指数法则:

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

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

数字相加:

2 + 2 = 4

= u^{4}

2 u^{2} - 1 = u^{4}

2 u^{2} - 1 = u^{4}

解

2 u^{2} - 1 = u^{4} : u = 1, u = - 1

2 u^{2} - 1 = u^{4}

交换两边

u^{4} = 2 u^{2} - 1

将

1

para o lado esquerdo

u^{4} = 2 u^{2} - 1

两边加上

1

u^{4} + 1 = 2 u^{2} - 1 + 1

化简

u^{4} + 1 = 2 u^{2}

u^{4} + 1 = 2 u^{2}

将

2 u^{2}

para o lado esquerdo

u^{4} + 1 = 2 u^{2}

两边减去

2 u^{2}

u^{4} + 1 - 2 u^{2} = 2 u^{2} - 2 u^{2}

化简

u^{4} + 1 - 2 u^{2} = 0

u^{4} + 1 - 2 u^{2} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 2 u^{2} + 1 = 0

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

v^{2} - 2 v + 1 = 0

解

v^{2} - 2 v + 1 = 0 : v = 1

v^{2} - 2 v + 1 = 0

使用求根公式求解

v^{2} - 2 v + 1 = 0

二次方程求根公式:

若

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

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

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 2^{2} - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

数字相减:

4 - 4 = 0

= 0

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

v = \frac{- ( - 2 )}{2 \cdot 1}

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

v = 1

二次方程组的解是:

v = 1

v = 1

代回

v = u^{2}

，求解

u

解

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

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

解为

u = 1, u = - 1

u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

的分母，令其等于零

解

u^{2} = 0 : u = 0

u^{2} = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 1, u = - 1

u = sec (x)

代回

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

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

sec (x) = 1 : x = 2 πn

sec (x) = 1

sec (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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

sec (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}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

9(sin(x)-0.6pi)+8=0

9 (sin (x) - 0.6 π) + 8 = 0

solvefor t,v=100cos(50t+20)v

so l v e f or t, v = 100 cos (50 t + 2 0^{\circ}) v

cos(2x)=sin(pi/2-x)

cos (2 x) = sin (\frac{π}{2} - x)

sin (x) = sin (5 0^{\circ})

solvefor n,y=-sin(2((3pi)/4+pin))2

so l v e f or n, y = - sin (2 (\frac{3 π}{4} + πn)) 2