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受欢迎的三角函数 >

sec^2(2x)-1=0

解答

sec^{2} (2 x) - 1 = 0

解答

x = πn, x = \frac{π}{2} + πn

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

求解步骤

sec^{2} (2 x) - 1 = 0

用替代法求解

sec^{2} (2 x) - 1 = 0

令

: sec (2 x) = u

u^{2} - 1 = 0

u^{2} - 1 = 0 : u = 1, u = - 1

u^{2} - 1 = 0

将

1

到右边

u^{2} - 1 = 0

两边加上

1

u^{2} - 1 + 1 = 0 + 1

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = sec (2 x)

代回

sec (2 x) = 1, sec (2 x) = - 1

sec (2 x) = 1, sec (2 x) = - 1

sec (2 x) = 1 : x = πn

sec (2 x) = 1

sec (2 x) = 1

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

解

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

两边除以

2

2 x = 2 πn

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2}

化简

x = πn

x = πn

x = πn

sec (2 x) = - 1 : x = \frac{π}{2} + πn

sec (2 x) = - 1

sec (2 x) = - 1

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

2 x = π + 2 πn

2 x = π + 2 πn

解

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

两边除以

2

2 x = π + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

合并所有解

x = πn, x = \frac{π}{2} + πn

作图

流行的例子

1 = 2 cos (x)

solvefor x,y=arcsin(x/(11))

so l v e f or x, y = arcsin (\frac{x}{11})

(sin(48))/(13)=(sin(b))/(16)

\frac{sin ( 4 8 ^{\circ} )}{13} = \frac{sin ( b )}{16}

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sqrt(8)cos(θ)+7=-6cos(θ)+sqrt(5)

8 cos (θ) + 7 = - 6 cos (θ) + 5