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受欢迎的三角函数 >

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

解答

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

解答

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

+1

度数

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

3

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

2 \cdot \frac{1}{sec ( x )}

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

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

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

用替代法求解

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

令

: sec (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

\frac{2}{u} u : 2

\frac{2}{u} u

分式相乘:

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

= \frac{2 u}{u}

约分:

u

= 2

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

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = - 3, c = 2

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

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

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

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

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 3)^{2} = 3^{2}

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

数字相乘:

4 \cdot 1 \cdot 2 = 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 1}

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

3 + 1 = 4

= \frac{4}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

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

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

使用法则

- (- a) = a

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

数字相减:

3 - 1 = 2

= \frac{2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2}{2}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = 2, u = 1

u = 2, u = 1

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (x)

代回

sec (x) = 2, sec (x) = 1

sec (x) = 2, sec (x) = 1

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

sec (x) = 2

sec (x) = 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}

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

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

sec (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

合并所有解

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

作图

流行的例子

cos^{2} (2 x) = 0

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

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

sqrt(2)cos(x)-sin(2x)=0

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

\frac{2}{3} = cos (x)

cos(2x)+3cos(x)=1

cos (2 x) + 3 cos (x) = 1