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受欢迎的三角函数 >

sin(2x)=2tan(2x)

解答

sin (2 x) = 2 tan (2 x)

解答

x = πn, x = \frac{π}{2} + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

求解步骤

sin (2 x) = 2 tan (2 x)

两边减去

2 tan (2 x)

sin (2 x) - 2 tan (2 x) = 0

用 sin, cos 表示

sin (2 x) - 2 tan (2 x)

使用基本三角恒等式:

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

= sin (2 x) - 2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

化简

sin (2 x) - 2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{sin ( 2 x ) cos ( 2 x ) - 2 sin ( 2 x )}{cos ( 2 x )}

sin (2 x) - 2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

乘

2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{2 sin ( 2 x )}{cos ( 2 x )}

2 \cdot \frac{sin ( 2 x )}{cos ( 2 x )}

分式相乘:

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

= \frac{sin ( 2 x ) \cdot 2}{cos ( 2 x )}

= sin (2 x) - \frac{2 sin ( 2 x )}{cos ( 2 x )}

将项转换为分式:

sin (2 x) = \frac{sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )}

= \frac{sin ( 2 x ) cos ( 2 x )}{cos ( 2 x )} - \frac{sin ( 2 x ) \cdot 2}{cos ( 2 x )}

因为分母相等，所以合并分式:

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

= \frac{sin ( 2 x ) cos ( 2 x ) - sin ( 2 x ) \cdot 2}{cos ( 2 x )}

= \frac{sin ( 2 x ) cos ( 2 x ) - 2 sin ( 2 x )}{cos ( 2 x )}

\frac{- 2 sin ( 2 x ) + cos ( 2 x ) sin ( 2 x )}{cos ( 2 x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 2 sin (2 x) + cos (2 x) sin (2 x) = 0

分解

- 2 sin (2 x) + cos (2 x) sin (2 x) : sin (2 x) (cos (2 x) - 2)

- 2 sin (2 x) + cos (2 x) sin (2 x)

因式分解出通项

sin (2 x)

= sin (2 x) (- 2 + cos (2 x))

sin (2 x) (cos (2 x) - 2) = 0

分别求解每个部分

sin (2 x) = 0 or cos (2 x) - 2 = 0

sin (2 x) = 0 : x = πn, x = \frac{π}{2} + πn

sin (2 x) = 0

sin (2 x) = 0

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 x = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

解

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

两边除以

2

2 x = 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2}

化简

x = πn

x = πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

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

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

cos (2 x) - 2 = 0 :

无解

cos (2 x) - 2 = 0

将

2

到右边

cos (2 x) - 2 = 0

两边加上

2

cos (2 x) - 2 + 2 = 0 + 2

化简

cos (2 x) = 2

cos (2 x) = 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = πn, x = \frac{π}{2} + πn

作图

流行的例子

-2cos(2θ)+12=4sin(θ)+9

- 2 cos (2 θ) + 12 = 4 sin (θ) + 9

sec^2(x)+sec(x)-2=0

sec^{2} (x) + sec (x) - 2 = 0

(cot(x)-1)(2sin(x)+1)=0

(cot (x) - 1) (2 sin (x) + 1) = 0

solvefor y,x=tan(y)

so l v e f or y, x = tan (y)

tan (x) = - \frac{12}{5}