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受欢迎的三角函数 >

cos(θ)=sec(θ)

解答

cos (θ) = sec (θ)

解答

θ = 2 πn, θ = π + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (θ) = sec (θ)

两边减去

sec (θ)

cos (θ) - sec (θ) = 0

使用三角恒等式改写

cos (θ) - sec (θ)

使用基本三角恒等式:

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

= \frac{1}{sec ( θ )} - sec (θ)

\frac{1}{sec ( θ )} - sec (θ) = 0

用替代法求解

\frac{1}{sec ( θ )} - sec (θ) = 0

令

: sec (θ) = u

\frac{1}{u} - u = 0

\frac{1}{u} - u = 0 : u = 1, u = - 1

\frac{1}{u} - u = 0

在两边乘以

u

\frac{1}{u} - u = 0

在两边乘以

u

\frac{1}{u} u - uu = 0 \cdot u

化简

\frac{1}{u} u - uu = 0 \cdot u

化简

\frac{1}{u} u : 1

\frac{1}{u} u

分式相乘:

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

= \frac{1 \cdot u}{u}

约分:

u

= 1

化简

- uu : - u^{2}

- uu

使用指数法则:

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

uu = u^{1 + 1}

= - u^{1 + 1}

数字相加:

1 + 1 = 2

= - u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

1 - u^{2} = 0

1 - u^{2} = 0

1 - u^{2} = 0

解

1 - u^{2} = 0 : u = 1, u = - 1

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

1 - u^{2} - 1 = 0 - 1

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

\frac{- u ^{2}}{- 1} = \frac{- 1}{- 1}

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = 1, u = - 1

验证解

找到无定义的点（奇点）:

u = 0

取

\frac{1}{u} - 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 = 1, u = - 1

u = sec (θ)

代回

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1, sec (θ) = - 1

sec (θ) = 1 : θ = 2 πn

sec (θ) = 1

sec (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

sec (θ) = - 1 : θ = π + 2 πn

sec (θ) = - 1

sec (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

合并所有解

θ = 2 πn, θ = π + 2 πn

作图

流行的例子

4.45tan^2(θ)-23tan(θ)+16.15=0

4.45 tan^{2} (θ) - 23 tan (θ) + 16.15 = 0

arccos(x)=arcsin(1/3)+arccos(1/4)

arccos (x) = arcsin (\frac{1}{3}) + arccos (\frac{1}{4})

sin(2x)=cos(40)

sin (2 x) = cos (4 0^{\circ})

-15sin(x)-8cos(x)=10

- 15 sin (x) - 8 cos (x) = 10

-5sec^2(x)+20=0

- 5 sec^{2} (x) + 20 = 0