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受欢迎的三角函数 >

2sin^2(2θ)=1+sin(2θ)

解答

2 sin^{2} (2 θ) = 1 + sin (2 θ)

解答

θ = \frac{π}{4} + πn, θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

+1

度数

θ = 4 5^{\circ} + 18 0^{\circ} n, θ = 10 5^{\circ} + 18 0^{\circ} n, θ = 16 5^{\circ} + 18 0^{\circ} n

求解步骤

2 sin^{2} (2 θ) = 1 + sin (2 θ)

用替代法求解

2 sin^{2} (2 θ) = 1 + sin (2 θ)

令

: sin (2 θ) = u

2 u^{2} = 1 + u

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

2 u^{2} = 1 + u

将

u

para o lado esquerdo

2 u^{2} = 1 + u

两边减去

u

2 u^{2} - u = 1 + u - u

化简

2 u^{2} - u = 1

2 u^{2} - u = 1

将

1

para o lado esquerdo

2 u^{2} - u = 1

两边减去

1

2 u^{2} - u - 1 = 1 - 1

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

(- 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 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

u_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 2}

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

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

使用法则

- (- a) = a

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

数字相加:

1 + 3 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

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

\frac{- ( - 1 ) - 3}{2 \cdot 2}

使用法则

- (- a) = a

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

数字相减:

1 - 3 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

二次方程组的解是:

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

u = sin (2 θ)

代回

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

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

sin (2 θ) = 1 : θ = \frac{π}{4} + πn

sin (2 θ) = 1

sin (2 θ) = 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}

2 θ = \frac{π}{2} + 2 πn

2 θ = \frac{π}{2} + 2 πn

解

2 θ = \frac{π}{2} + 2 πn : θ = \frac{π}{4} + πn

2 θ = \frac{π}{2} + 2 πn

两边除以

2

2 θ = \frac{π}{2} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

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

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

sin (2 θ) = - \frac{1}{2} : θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

sin (2 θ) = - \frac{1}{2}

sin (2 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}

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

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

解

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

2 θ = \frac{7 π}{6} + 2 πn

两边除以

2

2 θ = \frac{7 π}{6} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

使用分式法则:

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

= \frac{7 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

解

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

2 θ = \frac{11 π}{6} + 2 πn

两边除以

2

2 θ = \frac{11 π}{6} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

使用分式法则:

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

= \frac{11 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

合并所有解

θ = \frac{π}{4} + πn, θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

作图

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