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受欢迎的三角函数 >

sqrt(2)cos(x-pi/4)-1=0

解答

2 cos (x - \frac{π}{4}) - 1 = 0

解答

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

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 36 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (x - \frac{π}{4}) - 1 = 0

将

1

到右边

2 cos (x - \frac{π}{4}) - 1 = 0

两边加上

1

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

化简

2 cos (x - \frac{π}{4}) = 1

2 cos (x - \frac{π}{4}) = 1

两边除以

2

2 cos (x - \frac{π}{4}) = 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( x - \frac{π}{4} )}{2} : cos (x - \frac{π}{4})

\frac{2 cos ( x - \frac{π}{4} )}{2}

约分:

2

= cos (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}

cos (x - \frac{π}{4}) = \frac{2}{2}

cos (x - \frac{π}{4}) = \frac{2}{2}

cos (x - \frac{π}{4}) = \frac{2}{2}

cos (x - \frac{π}{4}) = \frac{2}{2}

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

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

解

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

x - \frac{π}{4} = \frac{π}{4} + 2 πn

将

\frac{π}{4}

到右边

x - \frac{π}{4} = \frac{π}{4} + 2 πn

两边加上

\frac{π}{4}

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

化简

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

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

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

\frac{π}{4} + 2 πn + \frac{π}{4}

对同类项分组

= \frac{π}{4} + \frac{π}{4} + 2 πn

合并分式

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

使用法则

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

= \frac{π + π}{4}

同类项相加:

π + π = 2 π

= \frac{2 π}{4}

约分:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

解

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

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

将

\frac{π}{4}

到右边

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

两边加上

\frac{π}{4}

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

化简

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

化简

x - \frac{π}{4} + \frac{π}{4} : x

x - \frac{π}{4} + \frac{π}{4}

同类项相加:

- \frac{π}{4} + \frac{π}{4} = 0

= x

化简

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

\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} : 2 π

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 = 8 π

= \frac{8 π}{4}

数字相除:

\frac{8}{4} = 2

= 2 π

= 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

x = 2 πn + 2 π

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

作图

流行的例子

0 = 3 cos (x)

tan(pi/2-x)+tan(x)-1=0

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

cos (1000 t) = 1

3sin(θ/4)-3/2 =0,0<= θ<= 2pi

3 sin (\frac{θ}{4}) - \frac{3}{2} = 0, 0 \leq θ \leq 2 π

(1-tan^2(A))/(2tan(A))=cot(A)

\frac{1 - tan ^{2} ( A )}{2 tan ( A )} = cot (A)