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受欢迎的三角函数 >

4+4sin(θ)= 3/(1-sin(θ))

解答

4 + 4 sin (θ) = \frac{3}{1 - sin ( θ )}

解答

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

+1

度数

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

4 + 4 sin (θ) = \frac{3}{1 - sin ( θ )}

用替代法求解

4 + 4 sin (θ) = \frac{3}{1 - sin ( θ )}

令

: sin (θ) = u

4 + 4 u = \frac{3}{1 - u}

4 + 4 u = \frac{3}{1 - u} : u = - \frac{1}{2}, u = \frac{1}{2}

4 + 4 u = \frac{3}{1 - u}

在两边乘以

1 - u

4 + 4 u = \frac{3}{1 - u}

在两边乘以

1 - u

4 (1 - u) + 4 u (1 - u) = \frac{3}{1 - u} (1 - u)

化简

4 (1 - u) + 4 u (1 - u) = 3

4 (1 - u) + 4 u (1 - u) = 3

解

4 (1 - u) + 4 u (1 - u) = 3 : u = - \frac{1}{2}, u = \frac{1}{2}

4 (1 - u) + 4 u (1 - u) = 3

展开

4 (1 - u) + 4 u (1 - u) : 4 - 4 u^{2}

4 (1 - u) + 4 u (1 - u)

乘开

4 (1 - u) : 4 - 4 u

4 (1 - u)

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = u

= 4 \cdot 1 - 4 u

数字相乘:

4 \cdot 1 = 4

= 4 - 4 u

= 4 - 4 u + 4 u (1 - u)

乘开

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

4 u (1 - u)

使用分配律:

a (b - c) = ab - a c

a = 4 u, b = 1, c = u

= 4 u \cdot 1 - 4 uu

= 4 \cdot 1 \cdot u - 4 uu

化简

4 \cdot 1 \cdot u - 4 uu : 4 u - 4 u^{2}

4 \cdot 1 \cdot u - 4 uu

4 \cdot 1 \cdot u = 4 u

4 \cdot 1 \cdot u

数字相乘:

4 \cdot 1 = 4

= 4 u

4 uu = 4 u^{2}

4 uu

使用指数法则:

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

uu = u^{1 + 1}

= 4 u^{1 + 1}

数字相加:

1 + 1 = 2

= 4 u^{2}

= 4 u - 4 u^{2}

= 4 u - 4 u^{2}

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

同类项相加:

- 4 u + 4 u = 0

= 4 - 4 u^{2}

4 - 4 u^{2} = 3

将

3

para o lado esquerdo

4 - 4 u^{2} = 3

两边减去

3

4 - 4 u^{2} - 3 = 3 - 3

化简

- 4 u^{2} + 1 = 0

- 4 u^{2} + 1 = 0

使用求根公式求解

- 4 u^{2} + 1 = 0

二次方程求根公式:

若

a = - 4, b = 0, c = 1

u_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

u_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

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

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

使用法则

0^{a} = 0

0^{2} = 0

= 0 - 4 (- 4) \cdot 1

使用法则

- (- a) = a

= 0 + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 0 + 16

数字相加:

0 + 16 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- 0 \pm 4}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 0 + 4}{2 ( - 4 )}, u_{2} = \frac{- 0 - 4}{2 ( - 4 )}

u = \frac{- 0 + 4}{2 ( - 4 )} : - \frac{1}{2}

\frac{- 0 + 4}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 0 + 4}{- 2 \cdot 4}

数字相加/相减:

- 0 + 4 = 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = \frac{- 0 - 4}{2 ( - 4 )} : \frac{1}{2}

\frac{- 0 - 4}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 0 - 4}{- 2 \cdot 4}

数字相减:

- 0 - 4 = - 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

二次方程组的解是:

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

取

\frac{3}{1 - u}

的分母，令其等于零

解

1 - u = 0 : u = 1

1 - u = 0

将

1

到右边

1 - u = 0

两边减去

1

1 - u - 1 = 0 - 1

化简

- u = - 1

- u = - 1

两边除以

- 1

- u = - 1

两边除以

- 1

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

化简

u = 1

u = 1

以下点无定义

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 (θ)

代回

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

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

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

sin (θ) = \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}

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

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

合并所有解

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

作图

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