tan(θ)=(sqrt(3))/3 ,180<= θ<= 360

解答

tan (θ) = \frac{3}{3}, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 21 0^{\circ}

弧度

θ = \frac{7 π}{6}

求解步骤

tan (θ) = \frac{3}{3}, 18 0^{\circ} \leq θ \leq 36 0^{\circ}

tan (θ) = \frac{3}{3}

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}

θ = 3 0^{\circ} + 18 0^{\circ} n

θ = 3 0^{\circ} + 18 0^{\circ} n

在

18 0^{\circ} \leq θ \leq 36 0^{\circ}

范围内的解

θ = 21 0^{\circ}

7.5 = 5 sin (\frac{π}{2} (x - 2)) + 5

2 cos (θ) + 23 = 3

- 22 = - 4 sin (2 θ)

8 cos (x) = 4 cos^{2} (x) + 3

\frac{26}{sin ( 8 8 ^{\circ} )} = \frac{21}{sin ( c )}