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受欢迎的三角函数 >

((sin^2(x)+1))/(1-sin^2(x))-1=2

解答

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

解答

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

+1

度数

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

求解步骤

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

用替代法求解

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

令

: sin (x) = u

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

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

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

将

1

到右边

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

两边加上

1

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

化简

\frac{u ^{2} + 1}{1 - u ^{2}} = 3

\frac{u ^{2} + 1}{1 - u ^{2}} = 3

在两边乘以

1 - u^{2}

\frac{u ^{2} + 1}{1 - u ^{2}} = 3

在两边乘以

1 - u^{2}

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

化简

u^{2} + 1 = 3 (1 - u^{2})

u^{2} + 1 = 3 (1 - u^{2})

解

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

u^{2} + 1 = 3 (1 - u^{2})

展开

3 (1 - u^{2}) : 3 - 3 u^{2}

3 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = u^{2}

= 3 \cdot 1 - 3 u^{2}

数字相乘:

3 \cdot 1 = 3

= 3 - 3 u^{2}

u^{2} + 1 = 3 - 3 u^{2}

将

1

到右边

u^{2} + 1 = 3 - 3 u^{2}

两边减去

1

u^{2} + 1 - 1 = 3 - 3 u^{2} - 1

化简

u^{2} = - 3 u^{2} + 2

u^{2} = - 3 u^{2} + 2

将

3 u^{2}

para o lado esquerdo

u^{2} = - 3 u^{2} + 2

两边加上

3 u^{2}

u^{2} + 3 u^{2} = - 3 u^{2} + 2 + 3 u^{2}

化简

4 u^{2} = 2

4 u^{2} = 2

两边除以

4

4 u^{2} = 2

两边除以

4

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

化简

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

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

对于

x^{2} = f (a)

解为

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

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

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

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

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

的分母，令其等于零

解

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

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

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

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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

sin (x) = \frac{1}{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}

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

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

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{1}{2}

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

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

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

合并所有解

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

作图

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