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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: sin (x) = u

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

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

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

在两边乘以

2 - u

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

在两边乘以

2 - u

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

化简

1 = u (2 - u)

1 = u (2 - u)

解

1 = u (2 - u) : u = 1

1 = u (2 - u)

展开

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

u (2 - u)

使用分配律:

a (b - c) = ab - a c

a = u, b = 2, c = u

= u \cdot 2 - uu

= 2 u - uu

uu = u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

= 2 u - u^{2}

1 = 2 u - u^{2}

交换两边

2 u - u^{2} = 1

将

1

para o lado esquerdo

2 u - u^{2} = 1

两边减去

1

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

化简

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

= 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

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

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

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

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

去除括号:

(- a) = - a

= \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

u = 1

二次方程组的解是:

u = 1

u = 1

验证解

找到无定义的点（奇点）:

u = 2

取

\frac{1}{2 - u}

的分母，令其等于零

解

2 - u = 0 : u = 2

2 - u = 0

将

2

到右边

2 - u = 0

两边减去

2

2 - u - 2 = 0 - 2

化简

- u = - 2

- u = - 2

两边除以

- 1

- u = - 2

两边除以

- 1

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

化简

u = 2

u = 2

以下点无定义

u = 2

将不在定义域的点与解相综合:

u = 1

u = sin (x)

代回

sin (x) = 1

sin (x) = 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn

作图

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