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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn

+1

度数

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

求解步骤

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

两边减去

1 + cos (x)

2 sec (x) - 1 - cos (x) = 0

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

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

用替代法求解

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

令

: sec (x) = u

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

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

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

在两边乘以

u

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

在两边乘以

u

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

化简

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

化简

- 1 \cdot u : - u

- 1 \cdot u

乘以:

1 \cdot u = u

= - 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

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

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

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

解

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

数字相加:

1 + 8 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

1 + 3 = 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{4}{4}

使用法则

\frac{a}{a} = 1

= 1

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

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

使用法则

- (- a) = a

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

数字相减:

1 - 3 = - 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}

二次方程组的解是:

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

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

验证解

找到无定义的点（奇点）:

u = 0

取

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

u = sec (x)

代回

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

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

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

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

无解

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

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

无解

合并所有解

x = 2 πn

作图

流行的例子

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

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

8(1-sin^2(x))+2sin(x)-7=0

8 (1 - sin^{2} (x)) + 2 sin (x) - 7 = 0

cot (5 x) = 1

tan^2(x)-6tan(x)+1=0

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

tan^2(x)-2tan(x)=1

tan^{2} (x) - 2 tan (x) = 1