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受欢迎的三角函数 >

sqrt(3)tan(3x)+sqrt(3)=0,0<= x<= 180

解答

3 tan (3 x) + 3 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

解答

x = 4 5^{\circ}, x = 10 5^{\circ}, x = 16 5^{\circ}

+1

弧度

x = \frac{π}{4}, x = \frac{7 π}{12}, x = \frac{11 π}{12}

求解步骤

3 tan (3 x) + 3 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

将

3

到右边

3 tan (3 x) + 3 = 0

两边减去

3

3 tan (3 x) + 3 - 3 = 0 - 3

化简

3 tan (3 x) = - 3

3 tan (3 x) = - 3

两边除以

3

3 tan (3 x) = - 3

两边除以

3

\frac{3 tan ( 3 x )}{3} = \frac{- 3}{3}

化简

tan (3 x) = - 1

tan (3 x) = - 1

tan (3 x) = - 1

的通解

tan (x)

周期表（周期为

18 0^{\circ} n

）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

3 x = 13 5^{\circ} + 18 0^{\circ} n

3 x = 13 5^{\circ} + 18 0^{\circ} n

解

3 x = 13 5^{\circ} + 18 0^{\circ} n : x = 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

3 x = 13 5^{\circ} + 18 0^{\circ} n

两边除以

3

3 x = 13 5^{\circ} + 18 0^{\circ} n

两边除以

3

\frac{3 x}{3} = \frac{13 5 ^{\circ}}{3} + \frac{18 0 ^{\circ} n}{3}

化简

\frac{3 x}{3} = \frac{13 5 ^{\circ}}{3} + \frac{18 0 ^{\circ} n}{3}

化简

\frac{3 x}{3} : x

\frac{3 x}{3}

数字相除:

\frac{3}{3} = 1

= x

化简

\frac{13 5 ^{\circ}}{3} + \frac{18 0 ^{\circ} n}{3} : 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

\frac{13 5 ^{\circ}}{3} + \frac{18 0 ^{\circ} n}{3}

\frac{13 5 ^{\circ}}{3} = 4 5^{\circ}

\frac{13 5 ^{\circ}}{3}

使用分式法则:

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

= \frac{54 0 ^{\circ}}{4 \cdot 3}

数字相乘:

4 \cdot 3 = 12

= 4 5^{\circ}

约分:

3

= 4 5^{\circ}

= 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

x = 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

x = 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

x = 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

x = 4 5^{\circ} + \frac{18 0 ^{\circ} n}{3}

在

0 \leq x \leq 18 0^{\circ}

范围内的解

x = 4 5^{\circ}, x = 10 5^{\circ}, x = 16 5^{\circ}

作图

流行的例子

2tan(θ)+4=tan(θ)+5

(sin(42)}{6.25}=\frac{sin(x))/9

(tan(x)+cot(x))/(csc(x))=cos(x)

2sin(x/2)=sqrt(2)