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受欢迎的三角函数 >

tan^2(2θ)=tan(2θ)

解答

tan^{2} (2 θ) = tan (2 θ)

解答

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{πn}{2}

+1

度数

θ = 22. 5^{\circ} + 9 0^{\circ} n, θ = 0^{\circ} + 9 0^{\circ} n

求解步骤

tan^{2} (2 θ) = tan (2 θ)

用替代法求解

tan^{2} (2 θ) = tan (2 θ)

令

: tan (2 θ) = u

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

u = tan (2 θ)

代回

tan (2 θ) = 1, tan (2 θ) = 0

tan (2 θ) = 1, tan (2 θ) = 0

tan (2 θ) = 1 : θ = \frac{π}{8} + \frac{πn}{2}

tan (2 θ) = 1

tan (2 θ) = 1

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}{4} + πn

2 θ = \frac{π}{4} + πn

解

2 θ = \frac{π}{4} + πn : θ = \frac{π}{8} + \frac{πn}{2}

2 θ = \frac{π}{4} + πn

两边除以

2

2 θ = \frac{π}{4} + πn

两边除以

2

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

化简

\frac{2 θ}{2} = \frac{\frac{π}{4}}{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}{4}}{2} + \frac{πn}{2} : \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

θ = \frac{π}{8} + \frac{πn}{2}

tan (2 θ) = 0 : θ = \frac{πn}{2}

tan (2 θ) = 0

tan (2 θ) = 0

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 = 0 + πn

2 θ = 0 + πn

解

2 θ = 0 + πn : θ = \frac{πn}{2}

2 θ = 0 + πn

0 + πn = πn

2 θ = πn

两边除以

2

2 θ = πn

两边除以

2

\frac{2 θ}{2} = \frac{πn}{2}

化简

θ = \frac{πn}{2}

θ = \frac{πn}{2}

θ = \frac{πn}{2}

合并所有解

θ = \frac{π}{8} + \frac{πn}{2}, θ = \frac{πn}{2}

作图

流行的例子

2cos(2x)=2sin(x)

2 cos (2 x) = 2 sin (x)

-sin(θ)+1+cos(θ)=0

- sin (θ) + 1 + cos (θ) = 0

sin (x) = \frac{1}{10}

sqrt(2)sin(2x)=-1

2 sin (2 x) = - 1

csc (2 x - \frac{π}{3}) = 2