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受欢迎的三角函数 >

sqrt(3)sec(2θ)+2=0

解答

3 sec (2 θ) + 2 = 0

解答

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

+1

度数

θ = 7 5^{\circ} + 18 0^{\circ} n, θ = 10 5^{\circ} + 18 0^{\circ} n

求解步骤

3 sec (2 θ) + 2 = 0

将

2

到右边

3 sec (2 θ) + 2 = 0

两边减去

2

3 sec (2 θ) + 2 - 2 = 0 - 2

化简

3 sec (2 θ) = - 2

3 sec (2 θ) = - 2

两边除以

3

3 sec (2 θ) = - 2

两边除以

3

\frac{3 sec ( 2 θ )}{3} = \frac{- 2}{3}

化简

\frac{3 sec ( 2 θ )}{3} = \frac{- 2}{3}

化简

\frac{3 sec ( 2 θ )}{3} : sec (2 θ)

\frac{3 sec ( 2 θ )}{3}

约分:

3

= sec (2 θ)

化简

\frac{- 2}{3} : - \frac{2 3}{3}

\frac{- 2}{3}

使用分式法则:

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

= - \frac{2}{3}

- \frac{2}{3}

有理化:

- \frac{2 3}{3}

- \frac{2}{3}

乘以共轭根式

\frac{3}{3}

= - \frac{2 3}{3 3}

33 = 3

33

使用根式运算法则:

a a = a

33 = 3

= 3

= - \frac{2 3}{3}

= - \frac{2 3}{3}

sec (2 θ) = - \frac{2 3}{3}

sec (2 θ) = - \frac{2 3}{3}

sec (2 θ) = - \frac{2 3}{3}

sec (2 θ) = - \frac{2 3}{3}

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

2 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

2 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

解

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

2 θ = \frac{5 π}{6} + 2 πn

两边除以

2

2 θ = \frac{5 π}{6} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{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}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

使用分式法则:

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

= \frac{5 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

解

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

2 θ = \frac{7 π}{6} + 2 πn

两边除以

2

2 θ = \frac{7 π}{6} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{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{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

使用分式法则:

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

= \frac{7 π}{6 \cdot 2}

数字相乘:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

作图

流行的例子

4sin^2(x)+1=-4sin(x)

4 sin^{2} (x) + 1 = - 4 sin (x)

cos (3 x + 6 0^{\circ}) = 0

2sin^2(x)+(-2-sqrt(2))sin(x)+sqrt(2)=0

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

3-sqrt(3)tan(2x)=4-2sqrt(3)tan(2x)

3 - 3 tan (2 x) = 4 - 23 tan (2 x)

cos(θ)= 3/(sqrt(34))

cos (θ) = \frac{3}{34}