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受欢迎的三角函数 >

4tan(x)+2sin(x)cos(x)=0

解答

4 tan (x) + 2 sin (x) cos (x) = 0

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

4 tan (x) + 2 sin (x) cos (x) = 0

用 sin, cos 表示

4 tan (x) + 2 cos (x) sin (x)

使用基本三角恒等式:

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

= 4 \cdot \frac{sin ( x )}{cos ( x )} + 2 cos (x) sin (x)

化简

4 \cdot \frac{sin ( x )}{cos ( x )} + 2 cos (x) sin (x) : \frac{4 sin ( x ) + 2 cos ^{2} ( x ) sin ( x )}{cos ( x )}

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

乘

4 \cdot \frac{sin ( x )}{cos ( x )} : \frac{4 sin ( x )}{cos ( x )}

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

分式相乘:

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

= \frac{sin ( x ) \cdot 4}{cos ( x )}

= \frac{4 sin ( x )}{cos ( x )} + 2 cos (x) sin (x)

将项转换为分式:

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

= \frac{sin ( x ) \cdot 4}{cos ( x )} + \frac{2 cos ( x ) sin ( x ) cos ( x )}{cos ( 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 ( x ) \cdot 4 + 2 cos ( x ) sin ( x ) cos ( x )}{cos ( x )}

sin (x) \cdot 4 + 2 cos (x) sin (x) cos (x) = 4 sin (x) + 2 cos^{2} (x) sin (x)

sin (x) \cdot 4 + 2 cos (x) sin (x) cos (x)

2 cos (x) sin (x) cos (x) = 2 cos^{2} (x) sin (x)

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

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= 2 sin (x) cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 2 sin (x) cos^{2} (x)

= 4 sin (x) + 2 cos^{2} (x) sin (x)

= \frac{4 sin ( x ) + 2 cos ^{2} ( x ) sin ( x )}{cos ( x )}

= \frac{4 sin ( x ) + 2 cos ^{2} ( x ) sin ( x )}{cos ( x )}

\frac{4 sin ( x ) + 2 cos ^{2} ( x ) sin ( x )}{cos ( x )} = 0

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

4 sin (x) + 2 cos^{2} (x) sin (x) = 0

分解

4 sin (x) + 2 cos^{2} (x) sin (x) : 2 sin (x) (cos^{2} (x) + 2)

4 sin (x) + 2 cos^{2} (x) sin (x)

将

4

改写为

2 \cdot 2

= 2 \cdot 2 sin (x) + 2 sin (x) cos^{2} (x)

因式分解出通项

2 sin (x)

= 2 sin (x) (2 + cos^{2} (x))

2 sin (x) (cos^{2} (x) + 2) = 0

分别求解每个部分

sin (x) = 0 or cos^{2} (x) + 2 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

cos^{2} (x) + 2 = 0 :

无解

cos^{2} (x) + 2 = 0

用替代法求解

cos^{2} (x) + 2 = 0

令

: cos (x) = u

u^{2} + 2 = 0

u^{2} + 2 = 0 : u = 2 i, u = - 2 i

u^{2} + 2 = 0

将

2

到右边

u^{2} + 2 = 0

两边减去

2

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

化简

u^{2} = - 2

u^{2} = - 2

对于

x^{2} = f (a)

解为

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

u = - 2, u = - - 2

化简

- 2 : 2 i

- 2

使用根式运算法则:

- a = - 1 a

- 2 = - 1 2

= - 1 2

使用虚数运算法则:

- 1 = i

= 2 i

化简

- - 2 : - 2 i

- - 2

化简

- 2 : 2 i

- 2

使用根式运算法则:

- a = - 1 a

- 2 = - 1 2

= - 1 2

使用虚数运算法则:

- 1 = i

= 2 i

= - 2 i

u = 2 i, u = - 2 i

u = cos (x)

代回

cos (x) = 2 i, cos (x) = - 2 i

cos (x) = 2 i, cos (x) = - 2 i

cos (x) = 2 i :

无解

cos (x) = 2 i

无解

cos (x) = - 2 i :

无解

cos (x) = - 2 i

无解

合并所有解

无解

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

3sin(x)+6cos(x)=4

3 sin (x) + 6 cos (x) = 4

3 tan (2 x + 1) = 2

tan(2x)-2=3tan(x)

tan (2 x) - 2 = 3 tan (x)

sin (x - \frac{π}{6}) = 0

arccot(x-2)=arccot(x-1)+arccot(x)

arccot (x - 2) = arccot (x - 1) + arccot (x)