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(sin(2x)+cos(2x))^2=1

解答

(sin (2 x) + cos (2 x))^{2} = 1

解答

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

+1

度数

x = 9 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

(sin (2 x) + cos (2 x))^{2} = 1

两边减去

1

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

分解

(sin (2 x) + cos (2 x))^{2} - 1 : (sin (2 x) + cos (2 x) + 1) (sin (2 x) + cos (2 x) - 1)

(sin (2 x) + cos (2 x))^{2} - 1

将

1

改写为

1^{2}

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

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

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

= ((sin (2 x) + cos (2 x)) + 1) ((sin (2 x) + cos (2 x)) - 1)

整理后得

= (sin (2 x) + cos (2 x) + 1) (sin (2 x) + cos (2 x) - 1)

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

分别求解每个部分

sin (2 x) + cos (2 x) + 1 = 0 or sin (2 x) + cos (2 x) - 1 = 0

sin (2 x) + cos (2 x) + 1 = 0 : x = πn + \frac{π}{2}, x = πn + \frac{3 π}{4}

sin (2 x) + cos (2 x) + 1 = 0

使用三角恒等式改写

sin (2 x) + cos (2 x) + 1

sin (2 x) + cos (2 x) = 2 sin (2 x + \frac{π}{4})

sin (2 x) + cos (2 x)

改写为

= 2 (\frac{1}{2} sin (2 x) + \frac{1}{2} cos (2 x))

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 x) + sin (\frac{π}{4}) cos (2 x))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 x + \frac{π}{4})

= 1 + 2 sin (2 x + \frac{π}{4})

1 + 2 sin (2 x + \frac{π}{4}) = 0

将

1

到右边

1 + 2 sin (2 x + \frac{π}{4}) = 0

两边减去

1

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

化简

2 sin (2 x + \frac{π}{4}) = - 1

2 sin (2 x + \frac{π}{4}) = - 1

两边除以

2

2 sin (2 x + \frac{π}{4}) = - 1

两边除以

2

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

化简

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

化简

\frac{2 sin ( 2 x + \frac{π}{4} )}{2} : sin (2 x + \frac{π}{4})

\frac{2 sin ( 2 x + \frac{π}{4} )}{2}

约分:

2

= sin (2 x + \frac{π}{4})

化简

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

- \frac{1}{2}

有理化:

- \frac{2}{2}

- \frac{1}{2}

乘以共轭根式

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

sin (2 x + \frac{π}{4}) = - \frac{2}{2}

sin (2 x + \frac{π}{4}) = - \frac{2}{2}

sin (2 x + \frac{π}{4}) = - \frac{2}{2}

sin (2 x + \frac{π}{4}) = - \frac{2}{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}

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

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

解

2 x + \frac{π}{4} = \frac{5 π}{4} + 2 πn : x = πn + \frac{π}{2}

2 x + \frac{π}{4} = \frac{5 π}{4} + 2 πn

将

\frac{π}{4}

到右边

2 x + \frac{π}{4} = \frac{5 π}{4} + 2 πn

两边减去

\frac{π}{4}

2 x + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{4} + 2 πn - \frac{π}{4}

化简

2 x + \frac{π}{4} - \frac{π}{4} = \frac{5 π}{4} + 2 πn - \frac{π}{4}

化简

2 x + \frac{π}{4} - \frac{π}{4} : 2 x

2 x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= 2 x

化简

\frac{5 π}{4} + 2 πn - \frac{π}{4} : 2 πn + π

\frac{5 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{5 π}{4}

合并分式

- \frac{π}{4} + \frac{5 π}{4} : π

使用法则

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

= \frac{- π + 5 π}{4}

同类项相加:

- π + 5 π = 4 π

= \frac{4 π}{4}

数字相除:

\frac{4}{4} = 1

= π

= 2 πn + π

2 x = 2 πn + π

2 x = 2 πn + π

2 x = 2 πn + π

两边除以

2

2 x = 2 πn + π

两边除以

2

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

化简

x = πn + \frac{π}{2}

x = πn + \frac{π}{2}

解

2 x + \frac{π}{4} = \frac{7 π}{4} + 2 πn : x = πn + \frac{3 π}{4}

2 x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

将

\frac{π}{4}

到右边

2 x + \frac{π}{4} = \frac{7 π}{4} + 2 πn

两边减去

\frac{π}{4}

2 x + \frac{π}{4} - \frac{π}{4} = \frac{7 π}{4} + 2 πn - \frac{π}{4}

化简

2 x + \frac{π}{4} - \frac{π}{4} = \frac{7 π}{4} + 2 πn - \frac{π}{4}

化简

2 x + \frac{π}{4} - \frac{π}{4} : 2 x

2 x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= 2 x

化简

\frac{7 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{3 π}{2}

\frac{7 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{7 π}{4}

合并分式

- \frac{π}{4} + \frac{7 π}{4} : \frac{3 π}{2}

使用法则

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

= \frac{- π + 7 π}{4}

同类项相加:

- π + 7 π = 6 π

= \frac{6 π}{4}

约分:

2

= \frac{3 π}{2}

= 2 πn + \frac{3 π}{2}

2 x = 2 πn + \frac{3 π}{2}

2 x = 2 πn + \frac{3 π}{2}

2 x = 2 πn + \frac{3 π}{2}

两边除以

2

2 x = 2 πn + \frac{3 π}{2}

两边除以

2

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

化简

\frac{2 x}{2} = \frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2} : πn + \frac{3 π}{4}

\frac{2 πn}{2} + \frac{\frac{3 π}{2}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

= πn + \frac{3 π}{4}

x = πn + \frac{3 π}{4}

x = πn + \frac{3 π}{4}

x = πn + \frac{3 π}{4}

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

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

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

使用三角恒等式改写

sin (2 x) + cos (2 x) - 1

sin (2 x) + cos (2 x) = 2 sin (2 x + \frac{π}{4})

sin (2 x) + cos (2 x)

改写为

= 2 (\frac{1}{2} sin (2 x) + \frac{1}{2} cos (2 x))

使用以下普通恒等式:

cos (\frac{π}{4}) = \frac{1}{2}

使用以下普通恒等式:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 x) + sin (\frac{π}{4}) cos (2 x))

使用角和恒等式:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 x + \frac{π}{4})

= - 1 + 2 sin (2 x + \frac{π}{4})

- 1 + 2 sin (2 x + \frac{π}{4}) = 0

将

1

到右边

- 1 + 2 sin (2 x + \frac{π}{4}) = 0

两边加上

1

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

化简

2 sin (2 x + \frac{π}{4}) = 1

2 sin (2 x + \frac{π}{4}) = 1

两边除以

2

2 sin (2 x + \frac{π}{4}) = 1

两边除以

2

\frac{2 sin ( 2 x + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( 2 x + \frac{π}{4} )}{2} = \frac{1}{2}

化简

\frac{2 sin ( 2 x + \frac{π}{4} )}{2} : sin (2 x + \frac{π}{4})

\frac{2 sin ( 2 x + \frac{π}{4} )}{2}

约分:

2

= sin (2 x + \frac{π}{4})

化简

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

sin (2 x + \frac{π}{4}) = \frac{2}{2}

sin (2 x + \frac{π}{4}) = \frac{2}{2}

sin (2 x + \frac{π}{4}) = \frac{2}{2}

sin (2 x + \frac{π}{4}) = \frac{2}{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}

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

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

解

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

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

两边减去

\frac{π}{4}

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

化简

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

解

2 x + \frac{π}{4} = \frac{3 π}{4} + 2 πn : x = πn + \frac{π}{4}

2 x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

将

\frac{π}{4}

到右边

2 x + \frac{π}{4} = \frac{3 π}{4} + 2 πn

两边减去

\frac{π}{4}

2 x + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

2 x + \frac{π}{4} - \frac{π}{4} = \frac{3 π}{4} + 2 πn - \frac{π}{4}

化简

2 x + \frac{π}{4} - \frac{π}{4} : 2 x

2 x + \frac{π}{4} - \frac{π}{4}

同类项相加:

\frac{π}{4} - \frac{π}{4} = 0

= 2 x

化简

\frac{3 π}{4} + 2 πn - \frac{π}{4} : 2 πn + \frac{π}{2}

\frac{3 π}{4} + 2 πn - \frac{π}{4}

对同类项分组

= 2 πn - \frac{π}{4} + \frac{3 π}{4}

合并分式

- \frac{π}{4} + \frac{3 π}{4} : \frac{π}{2}

使用法则

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

= \frac{- π + 3 π}{4}

同类项相加:

- π + 3 π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= 2 πn + \frac{π}{2}

2 x = 2 πn + \frac{π}{2}

2 x = 2 πn + \frac{π}{2}

2 x = 2 πn + \frac{π}{2}

两边除以

2

2 x = 2 πn + \frac{π}{2}

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

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

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

\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}

= πn + \frac{π}{4}

x = πn + \frac{π}{4}

x = πn + \frac{π}{4}

x = πn + \frac{π}{4}

x = πn, x = πn + \frac{π}{4}

合并所有解

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

流行的例子

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

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

cos^2(x)-sin^2(x)= 1/2

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

2cos^2(θ)-cos(θ)=0

2 cos^{2} (θ) - cos (θ) = 0

3 tan^{2} (θ) - 1 = 0

sin(2θ-pi/2)=-1

sin (2 θ - \frac{π}{2}) = - 1