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

解答

tan (x) = - 1, 0 < x < 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

度数

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

求解步骤

tan (x) = - 1, 0 < x < 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 < x < 2 π

范围内的解

x = \frac{3 π}{4}, x = \frac{7 π}{4}

sec^{2} (x) - 2 = tan^{2} (x), 0 \leq x \leq 2 π

4 cos (x) - 3 cos (x) - 1 = 0

3 sin (6 0^{\circ} - \frac{3 x}{4}) = 5 sin (\frac{3 x}{4} - 3 0^{\circ})

\frac{sin ( θ )}{60} = \frac{sin ( 8 0 ^{\circ} )}{280.87}

5.63 = 11 cos (\frac{2 π}{T} 0.137)