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受欢迎的三角函数 >

2sin(x)tan(x)+5=4sec(x)

解答

2 sin (x) tan (x) + 5 = 4 sec (x)

解答

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

+1

度数

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

求解步骤

2 sin (x) tan (x) + 5 = 4 sec (x)

两边减去

4 sec (x)

2 sin (x) tan (x) + 5 - 4 sec (x) = 0

用 sin, cos 表示

2 sin (x) \frac{sin ( x )}{cos ( x )} + 5 - 4 \cdot \frac{1}{cos ( x )} = 0

化简

2 sin (x) \frac{sin ( x )}{cos ( x )} + 5 - 4 \cdot \frac{1}{cos ( x )} : \frac{2 sin ^{2} ( x ) - 4 + 5 cos ( x )}{cos ( x )}

2 sin (x) \frac{sin ( x )}{cos ( x )} + 5 - 4 \cdot \frac{1}{cos ( x )}

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

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

分式相乘:

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

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

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

sin (x) \cdot 2 sin (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= 2 sin^{2} (x)

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

4 \cdot \frac{1}{cos ( x )} = \frac{4}{cos ( x )}

4 \cdot \frac{1}{cos ( x )}

分式相乘:

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

= \frac{1 \cdot 4}{cos ( x )}

数字相乘:

1 \cdot 4 = 4

= \frac{4}{cos ( x )}

= \frac{2 sin ^{2} ( x )}{cos ( x )} + 5 - \frac{4}{cos ( x )}

合并分式

\frac{2 sin ^{2} ( x )}{cos ( x )} - \frac{4}{cos ( x )} : \frac{2 sin ^{2} ( x ) - 4}{cos ( x )}

使用法则

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

= \frac{2 sin ^{2} ( x ) - 4}{cos ( x )}

= \frac{2 sin ^{2} ( x ) - 4}{cos ( x )} + 5

将项转换为分式:

5 = \frac{5 cos ( x )}{cos ( x )}

= \frac{2 sin ^{2} ( x ) - 4}{cos ( x )} + \frac{5 cos ( x )}{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 sin ^{2} ( x ) - 4 + 5 cos ( x )}{cos ( x )}

\frac{2 sin ^{2} ( x ) - 4 + 5 cos ( x )}{cos ( x )} = 0

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

2 sin^{2} (x) - 4 + 5 cos (x) = 0

两边减去

5 cos (x)

2 sin^{2} (x) - 4 = - 5 cos (x)

两边进行平方

(2 sin^{2} (x) - 4)^{2} = (- 5 cos (x))^{2}

两边减去

(- 5 cos (x))^{2}

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x) = 0

分解

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x) : (2 sin^{2} (x) - 4 + 5 cos (x)) (2 sin^{2} (x) - 4 - 5 cos (x))

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x)

将

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x)

改写为

(2 sin^{2} (x) - 4)^{2} - (5 cos (x))^{2}

(2 sin^{2} (x) - 4)^{2} - 25 cos^{2} (x)

将

25

改写为

5^{2}

= (2 sin^{2} (x) - 4)^{2} - 5^{2} cos^{2} (x)

使用指数法则:

a^{m} b^{m} = (ab)^{m}

5^{2} cos^{2} (x) = (5 cos (x))^{2}

= (2 sin^{2} (x) - 4)^{2} - (5 cos (x))^{2}

= (2 sin^{2} (x) - 4)^{2} - (5 cos (x))^{2}

使用平方差公式:

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

(2 sin^{2} (x) - 4)^{2} - (5 cos (x))^{2} = ((2 sin^{2} (x) - 4) + 5 cos (x)) ((2 sin^{2} (x) - 4) - 5 cos (x))

= ((2 sin^{2} (x) - 4) + 5 cos (x)) ((2 sin^{2} (x) - 4) - 5 cos (x))

整理后得

= (2 sin^{2} (x) + 5 cos (x) - 4) (2 sin^{2} (x) - 5 cos (x) - 4)

(2 sin^{2} (x) - 4 + 5 cos (x)) (2 sin^{2} (x) - 4 - 5 cos (x)) = 0

分别求解每个部分

2 sin^{2} (x) - 4 + 5 cos (x) = 0 or 2 sin^{2} (x) - 4 - 5 cos (x) = 0

2 sin^{2} (x) - 4 + 5 cos (x) = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

2 sin^{2} (x) - 4 + 5 cos (x) = 0

使用三角恒等式改写

- 4 + 2 sin^{2} (x) + 5 cos (x)

使用毕达哥拉斯恒等式:

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

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

= - 4 + 2 (1 - cos^{2} (x)) + 5 cos (x)

化简

- 4 + 2 (1 - cos^{2} (x)) + 5 cos (x) : 5 cos (x) - 2 cos^{2} (x) - 2

- 4 + 2 (1 - cos^{2} (x)) + 5 cos (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

2 \cdot 1 = 2

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

= - 4 + 2 - 2 cos^{2} (x) + 5 cos (x)

数字相加/相减:

- 4 + 2 = - 2

= 5 cos (x) - 2 cos^{2} (x) - 2

= 5 cos (x) - 2 cos^{2} (x) - 2

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

用替代法求解

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

令

: cos (x) = u

- 2 - 2 u^{2} + 5 u = 0

- 2 - 2 u^{2} + 5 u = 0 : u = \frac{1}{2}, u = 2

- 2 - 2 u^{2} + 5 u = 0

改写成标准形式

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

- 2 u^{2} + 5 u - 2 = 0

使用求根公式求解

- 2 u^{2} + 5 u - 2 = 0

二次方程求根公式:

若

a = - 2, b = 5, c = - 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 2 ) ( - 2 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 2 ) ( - 2 )}{2 ( - 2 )}

5^{2} - 4 (- 2) (- 2) = 3

5^{2} - 4 (- 2) (- 2)

使用法则

- (- a) = a

= 5^{2} - 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 5 \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- 5 + 3}{2 ( - 2 )}, u_{2} = \frac{- 5 - 3}{2 ( - 2 )}

u = \frac{- 5 + 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- 5 + 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 5 + 3}{- 2 \cdot 2}

数字相加/相减:

- 5 + 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{- 5 - 3}{2 ( - 2 )} : 2

\frac{- 5 - 3}{2 ( - 2 )}

去除括号:

(- a) = - a

= \frac{- 5 - 3}{- 2 \cdot 2}

数字相减:

- 5 - 3 = - 8

= \frac{- 8}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 8}{- 4}

使用分式法则:

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

= \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= 2

二次方程组的解是:

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

u = cos (x)

代回

cos (x) = \frac{1}{2}, cos (x) = 2

cos (x) = \frac{1}{2}, cos (x) = 2

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

的通解

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

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

cos (x) = 2 :

无解

cos (x) = 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

2 sin^{2} (x) - 4 - 5 cos (x) = 0 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

2 sin^{2} (x) - 4 - 5 cos (x) = 0

使用三角恒等式改写

- 4 + 2 sin^{2} (x) - 5 cos (x)

使用毕达哥拉斯恒等式:

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

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

= - 4 + 2 (1 - cos^{2} (x)) - 5 cos (x)

化简

- 4 + 2 (1 - cos^{2} (x)) - 5 cos (x) : - 2 cos^{2} (x) - 5 cos (x) - 2

- 4 + 2 (1 - cos^{2} (x)) - 5 cos (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

2 \cdot 1 = 2

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

= - 4 + 2 - 2 cos^{2} (x) - 5 cos (x)

数字相加/相减:

- 4 + 2 = - 2

= - 2 cos^{2} (x) - 5 cos (x) - 2

= - 2 cos^{2} (x) - 5 cos (x) - 2

- 2 - 2 cos^{2} (x) - 5 cos (x) = 0

用替代法求解

- 2 - 2 cos^{2} (x) - 5 cos (x) = 0

令

: cos (x) = u

- 2 - 2 u^{2} - 5 u = 0

- 2 - 2 u^{2} - 5 u = 0 : u = - 2, u = - \frac{1}{2}

- 2 - 2 u^{2} - 5 u = 0

改写成标准形式

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

- 2 u^{2} - 5 u - 2 = 0

使用求根公式求解

- 2 u^{2} - 5 u - 2 = 0

二次方程求根公式:

若

a = - 2, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) ( - 2 )}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 2 ) ( - 2 )}{2 ( - 2 )}

(- 5)^{2} - 4 (- 2) (- 2) = 3

(- 5)^{2} - 4 (- 2) (- 2)

使用法则

- (- a) = a

= (- 5)^{2} - 4 \cdot 2 \cdot 2

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 2 \cdot 2

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数字相减:

25 - 16 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 5 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 5 ) + 3}{2 ( - 2 )} : - 2

\frac{- ( - 5 ) + 3}{2 ( - 2 )}

去除括号:

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

= \frac{5 + 3}{- 2 \cdot 2}

数字相加:

5 + 3 = 8

= \frac{8}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{- 4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 5 ) - 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- ( - 5 ) - 3}{2 ( - 2 )}

去除括号:

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

= \frac{5 - 3}{- 2 \cdot 2}

数字相减:

5 - 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 = - 2, u = - \frac{1}{2}

u = cos (x)

代回

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

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

cos (x) = - 2 :

无解

cos (x) = - 2

- 1 \leq cos (x) \leq 1

无解

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

的通解

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

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

将解代入原方程进行验证

将它们代入

2 sin (x) tan (x) + 5 = 4 sec (x)

检验解是否符合

去除与方程不符的解。

检验

\frac{π}{3} + 2 πn

的解:

真

\frac{π}{3} + 2 πn

代入

n = 1

\frac{π}{3} + 2 π 1

对于

2 sin (x) tan (x) + 5 = 4 sec (x)

代入

x = \frac{π}{3} + 2 π 1

2 sin (\frac{π}{3} + 2 π 1) tan (\frac{π}{3} + 2 π 1) + 5 = 4 sec (\frac{π}{3} + 2 π 1)

整理后得

8 = 8

\Rightarrow 真

检验

\frac{5 π}{3} + 2 πn

的解:

真

\frac{5 π}{3} + 2 πn

代入

n = 1

\frac{5 π}{3} + 2 π 1

对于

2 sin (x) tan (x) + 5 = 4 sec (x)

代入

x = \frac{5 π}{3} + 2 π 1

2 sin (\frac{5 π}{3} + 2 π 1) tan (\frac{5 π}{3} + 2 π 1) + 5 = 4 sec (\frac{5 π}{3} + 2 π 1)

整理后得

8 = 8

\Rightarrow 真

检验

\frac{2 π}{3} + 2 πn

的解:

假

\frac{2 π}{3} + 2 πn

代入

n = 1

\frac{2 π}{3} + 2 π 1

对于

2 sin (x) tan (x) + 5 = 4 sec (x)

代入

x = \frac{2 π}{3} + 2 π 1

2 sin (\frac{2 π}{3} + 2 π 1) tan (\frac{2 π}{3} + 2 π 1) + 5 = 4 sec (\frac{2 π}{3} + 2 π 1)

整理后得

2 = - 8

\Rightarrow 假

检验

\frac{4 π}{3} + 2 πn

的解:

假

\frac{4 π}{3} + 2 πn

代入

n = 1

\frac{4 π}{3} + 2 π 1

对于

2 sin (x) tan (x) + 5 = 4 sec (x)

代入

x = \frac{4 π}{3} + 2 π 1

2 sin (\frac{4 π}{3} + 2 π 1) tan (\frac{4 π}{3} + 2 π 1) + 5 = 4 sec (\frac{4 π}{3} + 2 π 1)

整理后得

2 = - 8

\Rightarrow 假

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

作图

流行的例子

cos^2(x)-cos(x)-2=0,0<= x<= 2pi

2sin(x)=sqrt(3),0<= x<2pi

sec^2(x)+0.5tan(x)=1

8sin(x)=cos(x)-3