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受欢迎的三角函数 >

8cos(θ)=2sec(θ)

解答

8 cos (θ) = 2 sec (θ)

解答

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

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

8 cos (θ) = 2 sec (θ)

两边减去

2 sec (θ)

8 cos (θ) - 2 sec (θ) = 0

使用三角恒等式改写

- 2 sec (θ) + 8 cos (θ)

使用基本三角恒等式:

cos (x) = \frac{1}{sec ( x )}

= - 2 sec (θ) + 8 \cdot \frac{1}{sec ( θ )}

8 \cdot \frac{1}{sec ( θ )} = \frac{8}{sec ( θ )}

8 \cdot \frac{1}{sec ( θ )}

分式相乘:

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

= \frac{1 \cdot 8}{sec ( θ )}

数字相乘:

1 \cdot 8 = 8

= \frac{8}{sec ( θ )}

= - 2 sec (θ) + \frac{8}{sec ( θ )}

\frac{8}{sec ( θ )} - 2 sec (θ) = 0

用替代法求解

\frac{8}{sec ( θ )} - 2 sec (θ) = 0

令

: sec (θ) = u

\frac{8}{u} - 2 u = 0

\frac{8}{u} - 2 u = 0 : u = 2, u = - 2

\frac{8}{u} - 2 u = 0

在两边乘以

u

\frac{8}{u} - 2 u = 0

在两边乘以

u

\frac{8}{u} u - 2 uu = 0 \cdot u

化简

\frac{8}{u} u - 2 uu = 0 \cdot u

化简

\frac{8}{u} u : 8

\frac{8}{u} u

分式相乘:

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

= \frac{8 u}{u}

约分:

u

= 8

化简

- 2 uu : - 2 u^{2}

- 2 uu

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= - 2 u^{1 + 1}

数字相加:

1 + 1 = 2

= - 2 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

8 - 2 u^{2} = 0

8 - 2 u^{2} = 0

8 - 2 u^{2} = 0

解

8 - 2 u^{2} = 0 : u = 2, u = - 2

8 - 2 u^{2} = 0

将

8

到右边

8 - 2 u^{2} = 0

两边减去

8

8 - 2 u^{2} - 8 = 0 - 8

化简

- 2 u^{2} = - 8

- 2 u^{2} = - 8

两边除以

- 2

- 2 u^{2} = - 8

两边除以

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 8}{- 2}

化简

\frac{- 2 u ^{2}}{- 2} = \frac{- 8}{- 2}

化简

\frac{- 2 u ^{2}}{- 2} : u^{2}

\frac{- 2 u ^{2}}{- 2}

使用分式法则:

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

= \frac{2 u ^{2}}{2}

数字相除:

\frac{2}{2} = 1

= u^{2}

化简

\frac{- 8}{- 2} : 4

\frac{- 8}{- 2}

使用分式法则:

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

= \frac{8}{2}

数字相除:

\frac{8}{2} = 4

= 4

u^{2} = 4

u^{2} = 4

u^{2} = 4

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 4, u = - 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

u = 2, u = - 2

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{8}{u} - 2 u

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = 2, u = - 2

u = sec (θ)

代回

sec (θ) = 2, sec (θ) = - 2

sec (θ) = 2, sec (θ) = - 2

sec (θ) = 2 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sec (θ) = 2

sec (θ) = 2

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}

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

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

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

sec (θ) = - 2

sec (θ) = - 2

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}

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

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

合并所有解

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

作图

流行的例子

2sin(x)-csc(x)=0

2 sin (x) - csc (x) = 0

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

sin (x) + sin (x) = 0

3 sec (x) - 5 = 0

5 cos (θ) - 5 = 0

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

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