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受欢迎的三角函数 >

|tan(x)|=1,0<= x<= 2pi

解答

∣ tan (x) ∣ = 1, 0 \leq x \leq 2 π

解答

x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = \frac{π}{4}, x = \frac{5 π}{4}

+1

度数

x = 13 5^{\circ}, x = 31 5^{\circ}, x = 4 5^{\circ}, x = 22 5^{\circ}

求解步骤

∣ tan (x) ∣ = 1, 0 \leq x \leq 2 π

用替代法求解

∣ tan (x) ∣ = 1

令

: tan (x) = u

∣ u ∣ = 1

∣ u ∣ = 1 : u = - 1 or u = 1

∣ u ∣ = 1

使用绝对值运算法则: 若

∣ u ∣ = a, a > 0

，则

u = a

or

u = - a

u = - 1 or u = 1

u = tan (x)

代回

tan (x) = - 1 or tan (x) = 1

tan (x) = - 1 or tan (x) = 1

tan (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π

tan (x) = - 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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

tan (x) = 1, 0 \leq x \leq 2 π

tan (x) = 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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

在

0 \leq x \leq 2 π

范围内的解

x = \frac{π}{4}, x = \frac{5 π}{4}

合并所有解

x = \frac{3 π}{4}, x = \frac{7 π}{4}, x = \frac{π}{4}, x = \frac{5 π}{4}

作图

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