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受欢迎的三角函数 >

4sin^2(2θ)+6=9

解答

4 sin^{2} (2 θ) + 6 = 9

解答

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

+1

度数

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n, θ = 12 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n

求解步骤

4 sin^{2} (2 θ) + 6 = 9

用替代法求解

4 sin^{2} (2 θ) + 6 = 9

令

: sin (2 θ) = u

4 u^{2} + 6 = 9

4 u^{2} + 6 = 9 : u = \frac{3}{2}, u = - \frac{3}{2}

4 u^{2} + 6 = 9

将

6

到右边

4 u^{2} + 6 = 9

两边减去

6

4 u^{2} + 6 - 6 = 9 - 6

化简

4 u^{2} = 3

4 u^{2} = 3

两边除以

4

4 u^{2} = 3

两边除以

4

\frac{4 u ^{2}}{4} = \frac{3}{4}

化简

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

化简

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

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

u = sin (2 θ)

代回

sin (2 θ) = \frac{3}{2}, sin (2 θ) = - \frac{3}{2}

sin (2 θ) = \frac{3}{2}, sin (2 θ) = - \frac{3}{2}

sin (2 θ) = \frac{3}{2} : θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

sin (2 θ) = \frac{3}{2}

sin (2 θ) = \frac{3}{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{π}{3} + 2 πn, 2 θ = \frac{2 π}{3} + 2 πn

2 θ = \frac{π}{3} + 2 πn, 2 θ = \frac{2 π}{3} + 2 πn

解

2 θ = \frac{π}{3} + 2 πn : θ = \frac{π}{6} + πn

2 θ = \frac{π}{3} + 2 πn

两边除以

2

2 θ = \frac{π}{3} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{π}{3}}{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}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

使用分式法则:

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

= \frac{π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

解

2 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{3} + πn

2 θ = \frac{2 π}{3} + 2 πn

两边除以

2

2 θ = \frac{2 π}{3} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{2 π}{3}}{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{2 π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{3} + πn

\frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

使用分式法则:

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

= \frac{2 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{2 π}{6}

约分:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

sin (2 θ) = - \frac{3}{2} : θ = \frac{2 π}{3} + πn, θ = \frac{5 π}{6} + πn

sin (2 θ) = - \frac{3}{2}

sin (2 θ) = - \frac{3}{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{4 π}{3} + 2 πn, 2 θ = \frac{5 π}{3} + 2 πn

2 θ = \frac{4 π}{3} + 2 πn, 2 θ = \frac{5 π}{3} + 2 πn

解

2 θ = \frac{4 π}{3} + 2 πn : θ = \frac{2 π}{3} + πn

2 θ = \frac{4 π}{3} + 2 πn

两边除以

2

2 θ = \frac{4 π}{3} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{4 π}{3}}{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{4 π}{3}}{2} + \frac{2 πn}{2} : \frac{2 π}{3} + πn

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

使用分式法则:

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

= \frac{4 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{4 π}{6}

约分:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

解

2 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{6} + πn

2 θ = \frac{5 π}{3} + 2 πn

两边除以

2

2 θ = \frac{5 π}{3} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{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{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

使用分式法则:

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

= \frac{5 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

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

合并所有解

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

作图

流行的例子

tan(a)= 1/(sqrt(3))

tan (a) = \frac{1}{3}

tan (x) = 0.649

(sin(x)+cos(x))/(cos(x)+1)=tan(x)

\frac{sin ( x ) + cos ( x )}{cos ( x ) + 1} = tan (x)

cos(x)-cos(2x)=1

cos (x) - cos (2 x) = 1

tan(θ)= 1/(sqrt(2))

tan (θ) = \frac{1}{2}