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受欢迎的三角函数 >

sin^2(x)=2sin(x)+3,0<= x<= 2pi

解答

sin^{2} (x) = 2 sin (x) + 3, 0 \leq x \leq 2 π

解答

x = \frac{3 π}{2}

+1

度数

x = 27 0^{\circ}

求解步骤

sin^{2} (x) = 2 sin (x) + 3, 0 \leq x \leq 2 π

用替代法求解

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

令

: sin (x) = u

u^{2} = 2 u + 3

u^{2} = 2 u + 3 : u = 3, u = - 1

u^{2} = 2 u + 3

将

3

para o lado esquerdo

u^{2} = 2 u + 3

两边减去

3

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

化简

u^{2} - 3 = 2 u

u^{2} - 3 = 2 u

将

2 u

para o lado esquerdo

u^{2} - 3 = 2 u

两边减去

2 u

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

化简

u^{2} - 3 - 2 u = 0

u^{2} - 3 - 2 u = 0

改写成标准形式

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

u^{2} - 2 u - 3 = 0

使用求根公式求解

u^{2} - 2 u - 3 = 0

二次方程求根公式:

若

a = 1, b = - 2, c = - 3

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

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4

(- 2)^{2} - 4 \cdot 1 \cdot (- 3)

使用法则

- (- a) = a

= (- 2)^{2} + 4 \cdot 1 \cdot 3

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

数字相加:

4 + 12 = 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{- ( - 2 ) \pm 4}{2 \cdot 1}

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

2 + 4 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

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

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

使用法则

- (- a) = a

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

数字相减:

2 - 4 = - 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 = 3, u = - 1

u = sin (x)

代回

sin (x) = 3, sin (x) = - 1

sin (x) = 3, sin (x) = - 1

sin (x) = 3, 0 \leq x \leq 2 π :

无解

sin (x) = 3, 0 \leq x \leq 2 π

- 1 \leq sin (x) \leq 1

无解

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

sin (x) = - 1, 0 \leq x \leq 2 π

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

在

0 \leq x \leq 2 π

范围内的解

x = \frac{3 π}{2}

合并所有解

x = \frac{3 π}{2}

作图

流行的例子

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