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受欢迎的三角函数 >

tan(2θ)=1,0<= θ<= 360

解答

tan (2 θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

解答

θ = 22. 5^{\circ}, θ = 112. 5^{\circ}, θ = 202. 5^{\circ}, θ = 292. 5^{\circ}

+1

弧度

θ = \frac{π}{8}, θ = \frac{5 π}{8}, θ = \frac{9 π}{8}, θ = \frac{13 π}{8}

求解步骤

tan (2 θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

tan (2 θ) = 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}

2 θ = 4 5^{\circ} + 18 0^{\circ} n

2 θ = 4 5^{\circ} + 18 0^{\circ} n

解

2 θ = 4 5^{\circ} + 18 0^{\circ} n : θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

2 θ = 4 5^{\circ} + 18 0^{\circ} n

两边除以

2

2 θ = 4 5^{\circ} + 18 0^{\circ} n

两边除以

2

\frac{2 θ}{2} = \frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} = \frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} : 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

\frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2}

\frac{4 5 ^{\circ}}{2} = 22. 5^{\circ}

\frac{4 5 ^{\circ}}{2}

使用分式法则:

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

= \frac{18 0 ^{\circ}}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= 22. 5^{\circ}

= 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

θ = 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

在

0 \leq θ \leq 36 0^{\circ}

范围内的解

θ = 22. 5^{\circ}, θ = 112. 5^{\circ}, θ = 202. 5^{\circ}, θ = 292. 5^{\circ}

作图

流行的例子

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

tan^{3} (x) - 2 tan (x) = tan (x)

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

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

sin(x+16)=cos(11x+2)

sin (x + 1 6^{\circ}) = cos (11 x + 2)

sin^4(x/3)+cos^4(x/3)= 5/8

sin^{4} (\frac{x}{3}) + cos^{4} (\frac{x}{3}) = \frac{5}{8}

- 2 sin (x) = 1