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受欢迎的三角函数 >

2sin^2(u)=1+,u

解答

2 sin^{2} (u) = 1 +, u

解答

u \in R 无解

求解步骤

2 sin^{2} (u) = 1 +, u

用替代法求解

2 sin^{2} (u) = 1

令

: sin (u) = v

2 v^{2} = 1

2 v^{2} = 1 : v = \frac{1}{2}, v = - \frac{1}{2}

2 v^{2} = 1

两边除以

2

2 v^{2} = 1

两边除以

2

\frac{2 v ^{2}}{2} = \frac{1}{2}

化简

v^{2} = \frac{1}{2}

v^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

v = \frac{1}{2}, v = - \frac{1}{2}

v = sin (u)

代回

sin (u) = \frac{1}{2}, sin (u) = - \frac{1}{2}

sin (u) = \frac{1}{2}, sin (u) = - \frac{1}{2}

sin (u) = \frac{1}{2}, u :

无解

sin (u) = \frac{1}{2}, u

sin (u) = \frac{1}{2}

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}

u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 πn

u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 πn

在

u

范围内的解

无解

sin (u) = - \frac{1}{2}, u :

无解

sin (u) = - \frac{1}{2}, u

sin (u) = - \frac{1}{2}

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}

u = \frac{5 π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

u = \frac{5 π}{4} + 2 πn, u = \frac{7 π}{4} + 2 πn

在

u

范围内的解

无解

合并所有解

u \in R 无解

作图

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