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受欢迎的三角函数 >

sqrt(2)cos(2θ)+1=0

解答

2 cos (2 θ) + 1 = 0

解答

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

+1

度数

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

求解步骤

2 cos (2 θ) + 1 = 0

将

1

到右边

2 cos (2 θ) + 1 = 0

两边减去

1

2 cos (2 θ) + 1 - 1 = 0 - 1

化简

2 cos (2 θ) = - 1

2 cos (2 θ) = - 1

两边除以

2

2 cos (2 θ) = - 1

两边除以

2

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

化简

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

化简

\frac{2 cos ( 2 θ )}{2} : cos (2 θ)

\frac{2 cos ( 2 θ )}{2}

约分:

2

= cos (2 θ)

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 θ) = - \frac{2}{2}

cos (2 θ) = - \frac{2}{2}

cos (2 θ) = - \frac{2}{2}

cos (2 θ) = - \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 θ = \frac{3 π}{4} + 2 πn, 2 θ = \frac{5 π}{4} + 2 πn

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

解

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

2 θ = \frac{3 π}{4} + 2 πn

两边除以

2

2 θ = \frac{3 π}{4} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

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

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

θ = \frac{3 π}{8} + πn

解

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

2 θ = \frac{5 π}{4} + 2 πn

两边除以

2

2 θ = \frac{5 π}{4} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

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

θ = \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

θ = \frac{5 π}{8} + πn

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

作图

流行的例子

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

(2 sin (x)) + 1 = 0

sin (x) + 2 = 3

3tan(x)sin(x)=-2sin(x)

3 tan (x) sin (x) = - 2 sin (x)

2sin^2(x)+1=3sin(x)

2 sin^{2} (x) + 1 = 3 sin (x)

sin (θ) = \frac{3}{8}