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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

tan (x) + sec (x)

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

用 sin, cos 表示

2 cos (x) - \frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )} = 0

化简

2 cos (x) - \frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )} : \frac{2 cos ^{2} ( x ) - sin ( x ) - 1}{cos ( x )}

2 cos (x) - \frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )}

合并分式

- \frac{sin ( x )}{cos ( x )} - \frac{1}{cos ( x )} : \frac{- sin ( x ) - 1}{cos ( x )}

使用法则

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

= \frac{- sin ( x ) - 1}{cos ( x )}

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

将项转换为分式:

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

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

因为分母相等，所以合并分式:

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

= \frac{2 cos ( x ) cos ( x ) - sin ( x ) - 1}{cos ( x )}

2 cos (x) cos (x) - sin (x) - 1 = 2 cos^{2} (x) - sin (x) - 1

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

2 cos (x) cos (x) = 2 cos^{2} (x)

2 cos (x) cos (x)

使用指数法则:

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

cos (x) cos (x) = cos^{1 + 1} (x)

= 2 cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 2 cos^{2} (x)

= 2 cos^{2} (x) - sin (x) - 1

= \frac{2 cos ^{2} ( x ) - sin ( x ) - 1}{cos ( x )}

\frac{2 cos ^{2} ( x ) - sin ( x ) - 1}{cos ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 cos^{2} (x) - sin (x) - 1 = 0

两边加上

sin (x)

2 cos^{2} (x) - 1 = sin (x)

两边进行平方

(2 cos^{2} (x) - 1)^{2} = sin^{2} (x)

两边减去

sin^{2} (x)

(2 cos^{2} (x) - 1)^{2} - sin^{2} (x) = 0

分解

(2 cos^{2} (x) - 1)^{2} - sin^{2} (x) : (2 cos^{2} (x) - 1 + sin (x)) (2 cos^{2} (x) - 1 - sin (x))

(2 cos^{2} (x) - 1)^{2} - sin^{2} (x)

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos^{2} (x) - 1)^{2} - sin^{2} (x) = ((2 cos^{2} (x) - 1) + sin (x)) ((2 cos^{2} (x) - 1) - sin (x))

= ((2 cos^{2} (x) - 1) + sin (x)) ((2 cos^{2} (x) - 1) - sin (x))

整理后得

= (2 cos^{2} (x) + sin (x) - 1) (2 cos^{2} (x) - sin (x) - 1)

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

分别求解每个部分

2 cos^{2} (x) - 1 + sin (x) = 0 or 2 cos^{2} (x) - 1 - sin (x) = 0

2 cos^{2} (x) - 1 + sin (x) = 0 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

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

使用三角恒等式改写

- 1 + sin (x) + 2 cos^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

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

化简

- 1 + sin (x) + 2 (1 - sin^{2} (x)) : sin (x) - 2 sin^{2} (x) + 1

- 1 + sin (x) + 2 (1 - sin^{2} (x))

乘开

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (x)

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

化简

- 1 + sin (x) + 2 - 2 sin^{2} (x) : sin (x) - 2 sin^{2} (x) + 1

- 1 + sin (x) + 2 - 2 sin^{2} (x)

对同类项分组

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

1 + u - 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 ( - 2 ) \cdot 1}{2 ( - 2 )}

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

使用法则

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 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 ( - 2 )}

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相加/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}

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

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

去除括号:

(- a) = - a

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

数字相减:

- 1 - 3 = - 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 = sin (x)

代回

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn

2 cos^{2} (x) - 1 - sin (x) = 0 : x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

2 cos^{2} (x) - 1 - sin (x) = 0

使用三角恒等式改写

- 1 - sin (x) + 2 cos^{2} (x)

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

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

化简

- 1 - sin (x) + 2 (1 - sin^{2} (x)) : - 2 sin^{2} (x) - sin (x) + 1

- 1 - sin (x) + 2 (1 - sin^{2} (x))

乘开

2 (1 - sin^{2} (x)) : 2 - 2 sin^{2} (x)

2 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (x)

= 2 \cdot 1 - 2 sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (x)

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

化简

- 1 - sin (x) + 2 - 2 sin^{2} (x) : - 2 sin^{2} (x) - sin (x) + 1

- 1 - sin (x) + 2 - 2 sin^{2} (x)

对同类项分组

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

数字相加/相减:

- 1 + 2 = 1

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

1 - u - 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 ( - 2 ) \cdot 1}{2 ( - 2 )}

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

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

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

将解分隔开

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

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

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

去除括号:

(- a) = - a, - (- 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}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

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

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

去除括号:

(- a) = - a, - (- 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 = sin (x)

代回

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

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

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

sin (x) = - 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

\frac{7 π}{6} + 2 πn

的解:

假

\frac{7 π}{6} + 2 πn

代入

n = 1

\frac{7 π}{6} + 2 π 1

对于

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

代入

x = \frac{7 π}{6} + 2 π 1

2 cos (\frac{7 π}{6} + 2 π 1) = tan (\frac{7 π}{6} + 2 π 1) + sec (\frac{7 π}{6} + 2 π 1)

整理后得

- 1.73205 \dots = - 0.57735 \dots

\Rightarrow 假

检验

\frac{11 π}{6} + 2 πn

的解:

假

\frac{11 π}{6} + 2 πn

代入

n = 1

\frac{11 π}{6} + 2 π 1

对于

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

代入

x = \frac{11 π}{6} + 2 π 1

2 cos (\frac{11 π}{6} + 2 π 1) = tan (\frac{11 π}{6} + 2 π 1) + sec (\frac{11 π}{6} + 2 π 1)

整理后得

1.73205 \dots = 0.57735 \dots

\Rightarrow 假

检验

\frac{π}{2} + 2 πn

的解:

假

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

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

代入

x = \frac{π}{2} + 2 π 1

2 cos (\frac{π}{2} + 2 π 1) = tan (\frac{π}{2} + 2 π 1) + sec (\frac{π}{2} + 2 π 1)

整理后得

0 = \infty

\Rightarrow 假

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

x = \frac{3 π}{2} + 2 π 1

2 cos (\frac{3 π}{2} + 2 π 1) = tan (\frac{3 π}{2} + 2 π 1) + sec (\frac{3 π}{2} + 2 π 1)

未定义

\Rightarrow 假

检验

\frac{π}{6} + 2 πn

的解:

真

\frac{π}{6} + 2 πn

代入

n = 1

\frac{π}{6} + 2 π 1

对于

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

代入

x = \frac{π}{6} + 2 π 1

2 cos (\frac{π}{6} + 2 π 1) = tan (\frac{π}{6} + 2 π 1) + sec (\frac{π}{6} + 2 π 1)

整理后得

1.73205 \dots = 1.73205 \dots

\Rightarrow 真

检验

\frac{5 π}{6} + 2 πn

的解:

真

\frac{5 π}{6} + 2 πn

代入

n = 1

\frac{5 π}{6} + 2 π 1

对于

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

代入

x = \frac{5 π}{6} + 2 π 1

2 cos (\frac{5 π}{6} + 2 π 1) = tan (\frac{5 π}{6} + 2 π 1) + sec (\frac{5 π}{6} + 2 π 1)

整理后得

- 1.73205 \dots = - 1.73205 \dots

\Rightarrow 真

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

流行的例子

sin^2(x)cos(x)= 2/(3pi)

sin^{2} (x) cos (x) = \frac{2}{3 π}

tan(φ)=-1/(sqrt(6))

tan (φ) = - \frac{1}{6}

sin(θ)= 4/5 cos(θ)

sin (θ) = \frac{4}{5} cos (θ)

19= 1/2*7.9*6.2sin(x)

19 = \frac{1}{2} \cdot 7.9 \cdot 6.2 sin (x)

2tan(60-x)=tan(x)

2 tan (60 - x) = tan (x)