cos(t)=(sqrt(3))/2

解答

cos (t) = \frac{3}{2}

t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

度数

t = 3 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (t) = \frac{3}{2}

cos (t) = \frac{3}{2}

的通解

cos (x)

周期表（周期为

2 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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

t = \frac{π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

(cot (θ) + 1) (sin (θ) + 1) = 0

9 sec^{2} (θ) tan (θ) = 12 tan (θ)

3 tan (θ) = 1

2 cos (3 x) - 1 = 0, 0 \leq x \leq 2 π

16 cos^{2} (θ) = 4