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受欢迎的三角函数 >

4sin(θ)+4=(-3)/(sin(θ)-1)

解答

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

解答

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

+1

度数

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

求解步骤

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

用替代法求解

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

令

: sin (θ) = u

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

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

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

化简

\frac{- 3}{u - 1} : - \frac{3}{u - 1}

\frac{- 3}{u - 1}

使用分式法则:

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

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

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

在两边乘以

u - 1

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

在两边乘以

u - 1

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

化简

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

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

解

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

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

展开

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

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

乘开

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

4 u (u - 1)

使用分配律:

a (b - c) = ab - a c

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

= 4 uu - 4 u \cdot 1

= 4 uu - 4 \cdot 1 \cdot u

化简

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

4 uu - 4 \cdot 1 \cdot 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 \cdot 1 \cdot u = 4 u

4 \cdot 1 \cdot u

数字相乘:

4 \cdot 1 = 4

= 4 u

= 4 u^{2} - 4 u

= 4 u^{2} - 4 u

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

乘开

4 (u - 1) : 4 u - 4

4 (u - 1)

使用分配律:

a (b - c) = ab - a c

a = 4, b = u, c = 1

= 4 u - 4 \cdot 1

数字相乘:

4 \cdot 1 = 4

= 4 u - 4

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

同类项相加:

- 4 u + 4 u = 0

= 4 u^{2} - 4

4 u^{2} - 4 = - 3

将

3

para o lado esquerdo

4 u^{2} - 4 = - 3

两边加上

3

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

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

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

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

使用法则

0^{a} = 0

0^{2} = 0

= 0 - 4 \cdot 4 (- 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 \cdot 4}

将解分隔开

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

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

\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}

约分:

4

= \frac{1}{2}

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

\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}{u - 1}

的分母，令其等于零

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 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{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

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

合并所有解

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

作图

流行的例子

cos (θ) = 0.707

sin(x)+sin^2(x)+sin^3(x)+sin^4(x)=0

sin (x) + sin^{2} (x) + sin^{3} (x) + sin^{4} (x) = 0

sin(9x)-cos(x)=0

sin (9 x) - cos (x) = 0

cos(x)+2cos^2(x)=1

cos (x) + 2 cos^{2} (x) = 1

solvefor t,0.08=0.1cos(4t)

so l v e f or t, 0.08 = 0.1 cos (4 t)