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受欢迎的三角函数 >

sin^2(x)=|sin(x)|

解答

sin^{2} (x) = ∣ sin (x) ∣

解答

x = \frac{3 π}{2} + 2 πn, x = 2 πn, x = π + 2 πn, x = \frac{π}{2} + 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 = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) = ∣ sin (x) ∣

用替代法求解

sin^{2} (x) = ∣ sin (x) ∣

令

: sin (x) = u

u^{2} = ∣ u ∣

u^{2} = ∣ u ∣ : u = - 1 or u = 0 or u = 1

u^{2} = ∣ u ∣

找到正负区间

找到

∣ u ∣

的区间

u \geq 0

:

u \geq 0, ∣ u ∣ = u

对于

u \geq 0

改写

∣ u ∣ : ∣ u ∣ = u

使用绝对值运算法则: 若

u \geq 0

，则

∣ u ∣ = u

∣ u ∣ = u

u < 0

:

u < 0, ∣ u ∣ = - u

对于

u < 0

改写

∣ u ∣ : ∣ u ∣ = - u

使用绝对值运算法则: 若

u < 0

，则

∣ u ∣ = - u

∣ u ∣ = - u

确定区间:

u < 0, u \geq 0

∣ u ∣ u < 0 - u \geq 0 +

u < 0, u \geq 0

u < 0, u \geq 0

对每个区间解不等式

u < 0, u \geq 0

对于

u < 0 : u = - 1

对于

u < 0

改写

u^{2} = ∣ u ∣

为

u^{2} = - u

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

u^{2} = - u

将

u

para o lado esquerdo

u^{2} = - u

两边加上

u

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

化简

u^{2} + u = 0

u^{2} + u = 0

使用求根公式求解

u^{2} + u = 0

二次方程求根公式:

若

a = 1, b = 1, c = 0

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

使用法则

0 \cdot a = 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 1}

将解分隔开

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

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

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

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

= 0

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

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

数字相减:

- 1 - 1 = - 2

= \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

二次方程组的解是:

u = 0, u = - 1

合并区间

(u = - 1 or u = 0) and (u < 0)

合并重叠的区间

u = - 1 or u = 0 and u < 0

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 = - 1 or u = 0

and

u < 0

u = - 1

u = - 1

对于

u \geq 0 : u = 0 or u = 1

对于

u \geq 0

改写

u^{2} = ∣ u ∣

为

u^{2} = u

u^{2} = u : u = 1, u = 0

u^{2} = u

将

u

para o lado esquerdo

u^{2} = u

两边减去

u

u^{2} - u = u - u

化简

u^{2} - u = 0

u^{2} - u = 0

使用求根公式求解

u^{2} - u = 0

二次方程求根公式:

若

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

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

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

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

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

4 \cdot 1 \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 1}

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

1 + 1 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{0}{2}

使用法则

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

= 0

二次方程组的解是:

u = 1, u = 0

合并区间

(u = 0 or u = 1) and (u \geq 0)

合并重叠的区间

u = 0 or u = 1 and u \geq 0

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 = 1

and

u \geq 0

u = 0 or u = 1

u = 0 or u = 1

合并解:

u = - 1 or (u = 0 or u = 1)

u = - 1 or (u = 0 or u = 1)

u = - 1 or u = 0 or u = 1

u = sin (x)

代回

sin (x) = - 1 or sin (x) = 0 or sin (x) = 1

sin (x) = - 1 or sin (x) = 0 or sin (x) = 1

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

作图

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