sin(k)=0.5

解答

sin (k) = 0.5

k = \frac{π}{6} + 2 πn, k = \frac{5 π}{6} + 2 πn

度数

k = 3 0^{\circ} + 36 0^{\circ} n, k = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

sin (k) = 0.5

sin (k) = 0.5

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

k = \frac{π}{6} + 2 πn, k = \frac{5 π}{6} + 2 πn

k = \frac{π}{6} + 2 πn, k = \frac{5 π}{6} + 2 πn

cos (5 a + 4 0^{\circ}) = \frac{3}{5}

2 cos (x^{2}) - 2 = 0

so l v e f or x, sin (x) = - 2 \cdot cos (x)

cos (x) = \frac{1}{cos ^{2} ( x )}

2 cos (t) - 1 = 0, 0 <, = t \leq 2 π