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受欢迎的三角函数 >

2(sin(x)+1/2)^2+1=3|sin(x)+1/2 |

解答

2 (sin (x) + \frac{1}{2})^{2} + 1 = 3 sin (x) + \frac{1}{2}

解答

x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

2 (sin (x) + \frac{1}{2})^{2} + 1 = 3 sin (x) + \frac{1}{2}

用替代法求解

2 (sin (x) + \frac{1}{2})^{2} + 1 = 3 sin (x) + \frac{1}{2}

令

: sin (x) = u

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

2 (u + \frac{1}{2})^{2} + 1 = 3 u + \frac{1}{2} : u = - \frac{3}{2} or u = - 1 or u = 0 or u = \frac{1}{2}

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

找到正负区间

找到

u + \frac{1}{2}

的区间

u + \frac{1}{2} \geq 0

:

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

u + \frac{1}{2} \geq 0 : u \geq - \frac{1}{2}

u + \frac{1}{2} \geq 0

将

\frac{1}{2}

到右边

u + \frac{1}{2} \geq 0

两边减去

\frac{1}{2}

u + \frac{1}{2} - \frac{1}{2} \geq 0 - \frac{1}{2}

化简

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

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

对于

u + \frac{1}{2} \geq 0

改写

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

使用绝对值运算法则: 若

u \geq 0

，则

∣ u ∣ = u

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

u + \frac{1}{2} < 0

:

u < - \frac{1}{2}, u + \frac{1}{2} = - (u + \frac{1}{2})

u + \frac{1}{2} < 0 : u < - \frac{1}{2}

u + \frac{1}{2} < 0

将

\frac{1}{2}

到右边

u + \frac{1}{2} < 0

两边减去

\frac{1}{2}

u + \frac{1}{2} - \frac{1}{2} < 0 - \frac{1}{2}

化简

u < - \frac{1}{2}

u < - \frac{1}{2}

对于

u + \frac{1}{2} < 0

改写

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

使用绝对值运算法则: 若

u < 0

，则

∣ u ∣ = - u

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

确定区间:

u < - \frac{1}{2}, u \geq - \frac{1}{2}

u + \frac{1}{2} u < - \frac{1}{2} - u \geq - \frac{1}{2} +

u < - \frac{1}{2}, u \geq - \frac{1}{2}

u < - \frac{1}{2}, u \geq - \frac{1}{2}

对每个区间解不等式

u < - \frac{1}{2}, u \geq - \frac{1}{2}

对于

u < - \frac{1}{2} : u = - \frac{3}{2} or u = - 1

对于

u < - \frac{1}{2}

改写

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

为

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

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

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

展开

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

2 (u + \frac{1}{2})^{2} + 1

(u + \frac{1}{2})^{2} = u^{2} + u + \frac{1}{4}

(u + \frac{1}{2})^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u, b = \frac{1}{2}

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

化简

u^{2} + 2 u \frac{1}{2} + (\frac{1}{2})^{2} : u^{2} + u + \frac{1}{4}

u^{2} + 2 u \frac{1}{2} + (\frac{1}{2})^{2}

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

2 u \frac{1}{2}

分式相乘:

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

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

约分:

2

= u \cdot 1

乘以:

u \cdot 1 = u

= u

(\frac{1}{2})^{2} = \frac{1}{4}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用法则

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4}

= u^{2} + u + \frac{1}{4}

= u^{2} + u + \frac{1}{4}

= 2 (u^{2} + u + \frac{1}{4}) + 1

乘开

2 (u^{2} + u + \frac{1}{4}) : 2 u^{2} + 2 u + \frac{1}{2}

2 (u^{2} + u + \frac{1}{4})

打开括号

= 2 u^{2} + 2 u + 2 \cdot \frac{1}{4}

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

2 \cdot \frac{1}{4}

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

合并分式

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= \frac{1 \cdot 2}{2} + \frac{1}{2}

因为分母相等，所以合并分式:

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

= \frac{1 \cdot 2 + 1}{2}

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

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

展开

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

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

去除括号:

(- a) = - a

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

使用分配律:

a (b + c) = ab + a c

a = - 3, b = u, c = \frac{1}{2}

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

使用加减运算法则

+ (- a) = - a

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

3 \cdot \frac{1}{2} = \frac{3}{2}

3 \cdot \frac{1}{2}

分式相乘:

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

= \frac{1 \cdot 3}{2}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{2}

= - 3 u - \frac{3}{2}

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

将

\frac{3}{2}

para o lado esquerdo

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

两边加上

\frac{3}{2}

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

化简

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

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

将

3 u

para o lado esquerdo

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

两边加上

3 u

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

化简

2 u^{2} + 5 u + 3 = 0

2 u^{2} + 5 u + 3 = 0

使用求根公式求解

2 u^{2} + 5 u + 3 = 0

二次方程求根公式:

若

a = 2, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

5^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 5 \pm 1}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 5 + 1}{2 \cdot 2}, u_{2} = \frac{- 5 - 1}{2 \cdot 2}

u = \frac{- 5 + 1}{2 \cdot 2} : - 1

\frac{- 5 + 1}{2 \cdot 2}

数字相加/相减:

- 5 + 1 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 5 - 1}{2 \cdot 2} : - \frac{3}{2}

\frac{- 5 - 1}{2 \cdot 2}

数字相减:

- 5 - 1 = - 6

= \frac{- 6}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 6}{4}

使用分式法则:

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

= - \frac{6}{4}

约分:

2

= - \frac{3}{2}

二次方程组的解是:

u = - 1, u = - \frac{3}{2}

合并区间

(u = - \frac{3}{2} or u = - 1) and (u < - \frac{1}{2})

合并重叠的区间

u = - \frac{3}{2} or u = - 1 and u < - \frac{1}{2}

两个区间的交集是指同时存在于这两个区间的数的集合

u = - \frac{3}{2} or u = - 1

and

u < - \frac{1}{2}

u = - \frac{3}{2} or u = - 1

u = - \frac{3}{2} or u = - 1

对于

u \geq - \frac{1}{2} : u = 0 or u = \frac{1}{2}

对于

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

改写

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

为

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

2 (u + \frac{1}{2})^{2} + 1 = 3 (u + \frac{1}{2}) : u = \frac{1}{2}, u = 0

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

展开

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

2 (u + \frac{1}{2})^{2} + 1

(u + \frac{1}{2})^{2} = u^{2} + u + \frac{1}{4}

(u + \frac{1}{2})^{2}

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u, b = \frac{1}{2}

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

化简

u^{2} + 2 u \frac{1}{2} + (\frac{1}{2})^{2} : u^{2} + u + \frac{1}{4}

u^{2} + 2 u \frac{1}{2} + (\frac{1}{2})^{2}

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

2 u \frac{1}{2}

分式相乘:

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

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

约分:

2

= u \cdot 1

乘以:

u \cdot 1 = u

= u

(\frac{1}{2})^{2} = \frac{1}{4}

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

使用指数法则:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

使用法则

1^{a} = 1

1^{2} = 1

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

2^{2} = 4

= \frac{1}{4}

= u^{2} + u + \frac{1}{4}

= u^{2} + u + \frac{1}{4}

= 2 (u^{2} + u + \frac{1}{4}) + 1

乘开

2 (u^{2} + u + \frac{1}{4}) : 2 u^{2} + 2 u + \frac{1}{2}

2 (u^{2} + u + \frac{1}{4})

打开括号

= 2 u^{2} + 2 u + 2 \cdot \frac{1}{4}

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

2 \cdot \frac{1}{4}

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

合并分式

1 + \frac{1}{2} : \frac{3}{2}

1 + \frac{1}{2}

将项转换为分式:

1 = \frac{1 \cdot 2}{2}

= \frac{1 \cdot 2}{2} + \frac{1}{2}

因为分母相等，所以合并分式:

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

= \frac{1 \cdot 2 + 1}{2}

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

数字相乘:

1 \cdot 2 = 2

= 2 + 1

数字相加:

2 + 1 = 3

= 3

= \frac{3}{2}

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

展开

3 (u + \frac{1}{2}) : 3 u + \frac{3}{2}

3 (u + \frac{1}{2})

使用分配律:

a (b + c) = ab + a c

a = 3, b = u, c = \frac{1}{2}

= 3 u + 3 \cdot \frac{1}{2}

3 \cdot \frac{1}{2} = \frac{3}{2}

3 \cdot \frac{1}{2}

分式相乘:

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

= \frac{1 \cdot 3}{2}

数字相乘:

1 \cdot 3 = 3

= \frac{3}{2}

= 3 u + \frac{3}{2}

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

将

\frac{3}{2}

para o lado esquerdo

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

两边减去

\frac{3}{2}

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

化简

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

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

将

3 u

para o lado esquerdo

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

两边减去

3 u

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

化简

2 u^{2} - u = 0

2 u^{2} - u = 0

使用求根公式求解

2 u^{2} - u = 0

二次方程求根公式:

若

a = 2, b = - 1, c = 0

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

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

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

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

使用法则

- (- a) = a

= \frac{1 + 1}{2 \cdot 2}

数字相加:

1 + 1 = 2

= \frac{2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{2}{4}

约分:

2

= \frac{1}{2}

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

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

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

二次方程组的解是:

u = \frac{1}{2}, u = 0

合并区间

(u = 0 or u = \frac{1}{2}) and (u \geq - \frac{1}{2})

合并重叠的区间

u = 0 or u = \frac{1}{2} and u \geq - \frac{1}{2}

两个区间的交集是指同时存在于这两个区间的数的集合

u = 0 or u = \frac{1}{2}

and

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

u = 0 or u = \frac{1}{2}

u = 0 or u = \frac{1}{2}

合并解:

(u = - \frac{3}{2} or u = - 1) or (u = 0 or u = \frac{1}{2})

(u = - \frac{3}{2} or u = - 1) or (u = 0 or u = \frac{1}{2})

u = - \frac{3}{2} or u = - 1 or u = 0 or u = \frac{1}{2}

u = sin (x)

代回

sin (x) = - \frac{3}{2} or sin (x) = - 1 or sin (x) = 0 or sin (x) = \frac{1}{2}

sin (x) = - \frac{3}{2} or sin (x) = - 1 or sin (x) = 0 or sin (x) = \frac{1}{2}

sin (x) = - \frac{3}{2} :

无解

sin (x) = - \frac{3}{2}

- 1 \leq sin (x) \leq 1

无解

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

x = \frac{3 π}{2} + 2 πn

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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

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

合并所有解

x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

sin (x) = - \frac{1}{3}

5 cot (x) + 5 = 0

tan (x) + 1 = 2

cot(x)sec(x)+cot(x)=0

cot (x) sec (x) + cot (x) = 0

sin(2x)=(-sqrt(3))/2

sin (2 x) = \frac{- 3}{2}