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受欢迎的三角函数 >

cos(x)+2sec(x)=-3

解答

cos (x) + 2 sec (x) = - 3

解答

x = π + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (x) + 2 sec (x) = - 3

两边减去

- 3

cos (x) + 2 sec (x) + 3 = 0

使用三角恒等式改写

3 + cos (x) + 2 sec (x)

使用基本三角恒等式:

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

= 3 + \frac{1}{sec ( x )} + 2 sec (x)

3 + \frac{1}{sec ( x )} + 2 sec (x) = 0

用替代法求解

3 + \frac{1}{sec ( x )} + 2 sec (x) = 0

令

: sec (x) = u

3 + \frac{1}{u} + 2 u = 0

3 + \frac{1}{u} + 2 u = 0 : u = - \frac{1}{2}, u = - 1

3 + \frac{1}{u} + 2 u = 0

在两边乘以

u

3 + \frac{1}{u} + 2 u = 0

在两边乘以

u

3 u + \frac{1}{u} u + 2 uu = 0 \cdot u

化简

3 u + \frac{1}{u} u + 2 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

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

3 u + 1 + 2 u^{2} = 0

3 u + 1 + 2 u^{2} = 0

3 u + 1 + 2 u^{2} = 0

解

3 u + 1 + 2 u^{2} = 0 : u = - \frac{1}{2}, u = - 1

3 u + 1 + 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} + 3 u + 1 = 0

使用求根公式求解

2 u^{2} + 3 u + 1 = 0

二次方程求根公式:

若

a = 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

u = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 3 + 1}{2 \cdot 2}

数字相加/相减:

- 3 + 1 = - 2

= \frac{- 2}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

\frac{- 3 - 1}{2 \cdot 2}

数字相减:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

3 + \frac{1}{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 = - \frac{1}{2}, u = - 1

u = sec (x)

代回

sec (x) = - \frac{1}{2}, sec (x) = - 1

sec (x) = - \frac{1}{2}, sec (x) = - 1

sec (x) = - \frac{1}{2} :

无解

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

sec (x) \leq - 1 or sec (x) \geq 1

无解

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

sec (x) = - 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}

x = π + 2 πn

x = π + 2 πn

合并所有解

x = π + 2 πn

作图

流行的例子

0 = \frac{1}{5} sin (5 t)

tan^2(x)-1=0,0<= x<2pi

tan^{2} (x) - 1 = 0, 0 \leq x < 2 π

6sin^2(x)-5sin(x)+1=0

6 sin^{2} (x) - 5 sin (x) + 1 = 0

cos(θ)+1=sin(θ)

cos (θ) + 1 = sin (θ)

2 sec^{2} (x) - 2 = 0