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受欢迎的三角函数 >

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

解答

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

解答

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

求解步骤

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

用替代法求解

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

令

: sin (x) = u

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

\frac{2 - 2 u}{a} = \frac{2}{u} : u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

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

使用分式交叉相乘

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

使用分式交叉相乘: 若

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

则

a \cdot d = b \cdot c

(2 - 2 u) u = a \cdot 2

(2 - 2 u) u = a \cdot 2

解

(2 - 2 u) u = a \cdot 2 : u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}

(2 - 2 u) u = a \cdot 2

展开

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

(2 - 2 u) u

= u (2 - 2 u)

使用分配律:

a (b - c) = ab - a c

a = u, b = 2, c = 2 u

= u \cdot 2 - u \cdot 2 u

= 2 u - 2 uu

2 uu = 2 u^{2}

2 uu

使用指数法则:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= 2 u^{2}

= 2 u - 2 u^{2}

2 u - 2 u^{2} = a \cdot 2

将

a 2

para o lado esquerdo

2 u - 2 u^{2} = a \cdot 2

两边减去

a 2

2 u - 2 u^{2} - a \cdot 2 = a \cdot 2 - a \cdot 2

化简

2 u - 2 u^{2} - a \cdot 2 = 0

2 u - 2 u^{2} - a \cdot 2 = 0

改写成标准形式

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

- 2 u^{2} + 2 u - a \cdot 2 = 0

使用求根公式求解

- 2 u^{2} + 2 u - a \cdot 2 = 0

二次方程求根公式:

若

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

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

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

化简

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

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 2^{2} - 16 a

分解

2^{2} - 16 a : 4 (1 - 4 a)

2^{2} - 16 a

改写为

= 4 \cdot 1 - 4 \cdot 4 a

因式分解出通项

4

= 4 (1 - 4 a)

= 4 (1 - 4 a)

使用根式运算法则:

n ab = n a n b,

假定

a \geq 0, b \geq 0

= 4 - 4 a + 1

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= 2 - 4 a + 1

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

将解分隔开

u_{1} = \frac{- 2 + 2 1 - 4 a}{2 ( - 2 )}, u_{2} = \frac{- 2 - 2 1 - 4 a}{2 ( - 2 )}

u = \frac{- 2 + 2 1 - 4 a}{2 ( - 2 )} : - \frac{- 1 + - 4 a + 1}{2}

\frac{- 2 + 2 1 - 4 a}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2 1 - 4 a}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 + 2 - 4 a + 1}{- 4}

使用分式法则:

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

= - \frac{- 2 + 2 1 - 4 a}{4}

消掉

\frac{- 2 + 2 1 - 4 a}{4} : \frac{- 4 a + 1 - 1}{2}

\frac{- 2 + 2 1 - 4 a}{4}

分解

- 2 + 2 1 - 4 a : 2 (- 1 + 1 - 4 a)

- 2 + 2 1 - 4 a

改写为

= - 2 \cdot 1 + 2 1 - 4 a

因式分解出通项

2

= 2 (- 1 + 1 - 4 a)

= \frac{2 ( - 1 + 1 - 4 a )}{4}

约分:

2

= \frac{- 1 + - 4 a + 1}{2}

= - \frac{- 4 a + 1 - 1}{2}

= - \frac{- 1 + - 4 a + 1}{2}

u = \frac{- 2 - 2 1 - 4 a}{2 ( - 2 )} : \frac{- 4 a + 1 + 1}{2}

\frac{- 2 - 2 1 - 4 a}{2 ( - 2 )}

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2 - 2 - 4 a + 1}{- 4}

使用分式法则:

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

= - \frac{- 2 - 2 1 - 4 a}{4}

消掉

\frac{- 2 - 2 1 - 4 a}{4} : - \frac{- 4 a + 1 + 1}{2}

\frac{- 2 - 2 1 - 4 a}{4}

分解

- 2 - 2 1 - 4 a : - 2 (1 + 1 - 4 a)

- 2 - 2 1 - 4 a

改写为

= - 2 \cdot 1 - 2 1 - 4 a

因式分解出通项

2

= - 2 (1 + 1 - 4 a)

= - \frac{2 ( 1 + 1 - 4 a )}{4}

约分:

2

= - \frac{- 4 a + 1 + 1}{2}

= - (- \frac{- 4 a + 1 + 1}{2})

使用法则

- (- a) = a

= \frac{- 4 a + 1 + 1}{2}

二次方程组的解是:

u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}

u = - \frac{- 1 + - 4 a + 1}{2}, u = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

u = sin (x)

代回

sin (x) = - \frac{- 1 + - 4 a + 1}{2}, sin (x) = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

sin (x) = - \frac{- 1 + - 4 a + 1}{2}, sin (x) = \frac{- 4 a + 1 + 1}{2}; a \neq = 0

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

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

使用反三角函数性质

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

sin (x) = - \frac{- 1 + - 4 a + 1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn

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

sin (x) = \frac{- 4 a + 1 + 1}{2}

使用反三角函数性质

sin (x) = \frac{- 4 a + 1 + 1}{2}

sin (x) = \frac{- 4 a + 1 + 1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

合并所有解

x = arcsin (- \frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = π + arcsin (\frac{- 1 + - 4 a + 1}{2}) + 2 πn, x = arcsin (\frac{- 4 a + 1 + 1}{2}) + 2 πn, x = π + arcsin (- \frac{- 4 a + 1 + 1}{2}) + 2 πn

作图

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