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受欢迎的三角函数 >

tan^2(2θ)-3=0

解答

tan^{2} (2 θ) - 3 = 0

解答

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

+1

度数

θ = 3 0^{\circ} + 9 0^{\circ} n, θ = 6 0^{\circ} + 9 0^{\circ} n

求解步骤

tan^{2} (2 θ) - 3 = 0

用替代法求解

tan^{2} (2 θ) - 3 = 0

令

: tan (2 θ) = u

u^{2} - 3 = 0

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

u^{2} - 3 = 0

将

3

到右边

u^{2} - 3 = 0

两边加上

3

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

化简

u^{2} = 3

u^{2} = 3

对于

x^{2} = f (a)

解为

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

u = 3, u = - 3

u = tan (2 θ)

代回

tan (2 θ) = 3, tan (2 θ) = - 3

tan (2 θ) = 3, tan (2 θ) = - 3

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

tan (2 θ) = 3

tan (2 θ) = 3

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 θ = \frac{π}{3} + πn

2 θ = \frac{π}{3} + πn

解

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

2 θ = \frac{π}{3} + πn

两边除以

2

2 θ = \frac{π}{3} + πn

两边除以

2

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

化简

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

\frac{\frac{π}{3}}{2} + \frac{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{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

tan (2 θ) = - 3 : θ = \frac{π}{3} + \frac{πn}{2}

tan (2 θ) = - 3

tan (2 θ) = - 3

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 θ = \frac{2 π}{3} + πn

2 θ = \frac{2 π}{3} + πn

解

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

2 θ = \frac{2 π}{3} + πn

两边除以

2

2 θ = \frac{2 π}{3} + πn

两边除以

2

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

化简

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

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

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

合并所有解

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

作图

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