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受欢迎的三角函数 >

(1-sin(2x))*(1-cos^2(x))=0

解答

(1 - sin (2 x)) \cdot (1 - cos^{2} (x)) = 0

解答

x = \frac{π}{4} + πn, x = 2 πn, x = π + 2 πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

(1 - sin (2 x)) (1 - cos^{2} (x)) = 0

分别求解每个部分

1 - sin (2 x) = 0 or 1 - cos^{2} (x) = 0

1 - sin (2 x) = 0 : x = \frac{π}{4} + πn

1 - sin (2 x) = 0

将

1

到右边

1 - sin (2 x) = 0

两边减去

1

1 - sin (2 x) - 1 = 0 - 1

化简

- sin (2 x) = - 1

- sin (2 x) = - 1

两边除以

- 1

- sin (2 x) = - 1

两边除以

- 1

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

化简

sin (2 x) = 1

sin (2 x) = 1

sin (2 x) = 1

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}

2 x = \frac{π}{2} + 2 πn

2 x = \frac{π}{2} + 2 πn

解

2 x = \frac{π}{2} + 2 πn : x = \frac{π}{4} + πn

2 x = \frac{π}{2} + 2 πn

两边除以

2

2 x = \frac{π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{π}{2}}{2} + \frac{2 πn}{2} : \frac{π}{4} + πn

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

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

使用分式法则:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

1 - cos^{2} (x) = 0 : x = 2 πn, x = π + 2 πn

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

用替代法求解

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

令

: cos (x) = u

1 - u^{2} = 0

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

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

代回

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 2 πn

cos (x) = 1

cos (x) = 1

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}

x = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

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}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn

合并所有解

x = \frac{π}{4} + πn, x = 2 πn, x = π + 2 πn

作图

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