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受欢迎的三角函数 >

sqrt(2)cos(2x)=-1

解答

2 cos (2 x) = - 1

解答

x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn

+1

度数

x = 67. 5^{\circ} + 18 0^{\circ} n, x = 112. 5^{\circ} + 18 0^{\circ} n

求解步骤

2 cos (2 x) = - 1

两边除以

2

2 cos (2 x) = - 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( 2 x )}{2} : cos (2 x)

\frac{2 cos ( 2 x )}{2}

约分:

2

= cos (2 x)

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}

cos (2 x) = - \frac{2}{2}

cos (2 x) = - \frac{2}{2}

cos (2 x) = - \frac{2}{2}

cos (2 x) = - \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}

2 x = \frac{3 π}{4} + 2 πn, 2 x = \frac{5 π}{4} + 2 πn

2 x = \frac{3 π}{4} + 2 πn, 2 x = \frac{5 π}{4} + 2 πn

解

2 x = \frac{3 π}{4} + 2 πn : x = \frac{3 π}{8} + πn

2 x = \frac{3 π}{4} + 2 πn

两边除以

2

2 x = \frac{3 π}{4} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{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{3 π}{4}}{2} + \frac{2 πn}{2} : \frac{3 π}{8} + πn

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

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

使用分式法则:

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

= \frac{3 π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

解

2 x = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

2 x = \frac{5 π}{4} + 2 πn

两边除以

2

2 x = \frac{5 π}{4} + 2 πn

两边除以

2

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

化简

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{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{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

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

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

使用分式法则:

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

= \frac{5 π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn

作图

流行的例子

sec^2(a)cot(a)=8

sec^{2} (a) cot (a) = 8

6 cos (x) + 2 = 0

solvefor y,ln(sin^2(3x))=e^x+arccot(y)

so l v e f or y, ln (sin^{2} (3 x)) = e^{x} + arccot (y)

cos (A) = \frac{8}{15}

cot^2(θ)-1=csc^2(θ)

cot^{2} (θ) - 1 = csc^{2} (θ)