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受欢迎的三角函数 >

sqrt(3)*tan(x)+2sec(x)=1

解答

3 \cdot tan (x) + 2 sec (x) = 1

解答

x = \frac{5 π}{3} + 2 πn

+1

度数

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

求解步骤

3 tan (x) + 2 sec (x) = 1

两边减去

1

3 tan (x) + 2 sec (x) - 1 = 0

用 sin, cos 表示

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

化简

3 \frac{sin ( x )}{cos ( x )} + 2 \cdot \frac{1}{cos ( x )} - 1 : \frac{3 sin ( x ) + 2 - cos ( x )}{cos ( x )}

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

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

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

分式相乘:

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

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

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

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

分式相乘:

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

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

数字相乘:

1 \cdot 2 = 2

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

= \frac{3 sin ( x )}{cos ( x )} + \frac{2}{cos ( x )} - 1

合并分式

\frac{3 sin ( x )}{cos ( x )} + \frac{2}{cos ( x )} : \frac{3 sin ( x ) + 2}{cos ( x )}

使用法则

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

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

= \frac{3 sin ( x ) + 2}{cos ( x )} - 1

将项转换为分式:

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

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

乘以:

1 \cdot cos (x) = cos (x)

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

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

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

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

两边加上

cos (x)

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

两边进行平方

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

两边减去

cos^{2} (x)

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

使用三角恒等式改写

(2 + sin (x) 3)^{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)

= (2 + sin (x) 3)^{2} - (1 - sin^{2} (x))

化简

(2 + sin (x) 3)^{2} - (1 - sin^{2} (x)) : 4 sin^{2} (x) + 43 sin (x) + 3

(2 + sin (x) 3)^{2} - (1 - sin^{2} (x))

= (2 + 3 sin (x))^{2} - (1 - sin^{2} (x))

(2 + sin (x) 3)^{2} : 4 + 43 sin (x) + 3 sin^{2} (x)

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = 2, b = sin (x) 3

= 2^{2} + 2 \cdot 2 sin (x) 3 + (sin (x) 3)^{2}

化简

2^{2} + 2 \cdot 2 sin (x) 3 + (sin (x) 3)^{2} : 4 + 43 sin (x) + 3 sin^{2} (x)

2^{2} + 2 \cdot 2 sin (x) 3 + (sin (x) 3)^{2}

2^{2} = 4

2^{2}

2^{2} = 4

= 4

2 \cdot 2 sin (x) 3 = 43 sin (x)

2 \cdot 2 sin (x) 3

数字相乘:

2 \cdot 2 = 4

= 43 sin (x)

(sin (x) 3)^{2} = 3 sin^{2} (x)

(sin (x) 3)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= (3)^{2} sin^{2} (x)

(3)^{2} : 3

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 3

= sin^{2} (x) \cdot 3

= 4 + 43 sin (x) + 3 sin^{2} (x)

= 4 + 43 sin (x) + 3 sin^{2} (x)

= 4 + 43 sin (x) + 3 sin^{2} (x) - (1 - sin^{2} (x))

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

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

打开括号

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

使用加减运算法则

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

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

= 4 + 43 sin (x) + 3 sin^{2} (x) - 1 + sin^{2} (x)

化简

4 + 43 sin (x) + 3 sin^{2} (x) - 1 + sin^{2} (x) : 4 sin^{2} (x) + 43 sin (x) + 3

4 + 43 sin (x) + 3 sin^{2} (x) - 1 + sin^{2} (x)

对同类项分组

= 43 sin (x) + 3 sin^{2} (x) + sin^{2} (x) + 4 - 1

同类项相加:

3 sin^{2} (x) + sin^{2} (x) = 4 sin^{2} (x)

= 43 sin (x) + 4 sin^{2} (x) + 4 - 1

数字相加/相减:

4 - 1 = 3

= 4 sin^{2} (x) + 43 sin (x) + 3

= 4 sin^{2} (x) + 43 sin (x) + 3

= 4 sin^{2} (x) + 43 sin (x) + 3

3 + 4 sin^{2} (x) + 4 sin (x) 3 = 0

用替代法求解

3 + 4 sin^{2} (x) + 4 sin (x) 3 = 0

令

: sin (x) = u

3 + 4 u^{2} + 4 u 3 = 0

3 + 4 u^{2} + 4 u 3 = 0 : u = - \frac{3}{2}

3 + 4 u^{2} + 4 u 3 = 0

改写成标准形式

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

4 u^{2} + 43 u + 3 = 0

使用求根公式求解

4 u^{2} + 43 u + 3 = 0

二次方程求根公式:

若

a = 4, b = 43, c = 3

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

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

(43)^{2} - 4 \cdot 4 \cdot 3 = 0

(43)^{2} - 4 \cdot 4 \cdot 3

(43)^{2} = 4^{2} \cdot 3

(43)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

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

(3)^{2} : 3

使用根式运算法则:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分式相乘:

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

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

约分:

2

= 1

= 3

= 4^{2} \cdot 3

4 \cdot 4 \cdot 3 = 48

4 \cdot 4 \cdot 3

数字相乘:

4 \cdot 4 \cdot 3 = 48

= 48

= 4^{2} \cdot 3 - 48

4^{2} \cdot 3 = 48

4^{2} \cdot 3

4^{2} = 16

= 16 \cdot 3

数字相乘:

16 \cdot 3 = 48

= 48

= 48 - 48

数字相减:

48 - 48 = 0

= 0

u_{1, 2} = \frac{- 4 3 \pm 0}{2 \cdot 4}

u = \frac{- 4 3}{2 \cdot 4}

\frac{- 4 3}{2 \cdot 4} = - \frac{3}{2}

\frac{- 4 3}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 4 3}{8}

使用分式法则:

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

= - \frac{4 3}{8}

约分:

4

= - \frac{3}{2}

u = - \frac{3}{2}

二次方程组的解是:

u = - \frac{3}{2}

u = sin (x)

代回

sin (x) = - \frac{3}{2}

sin (x) = - \frac{3}{2}

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

sin (x) = - \frac{3}{2}

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

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

合并所有解

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

将解代入原方程进行验证

将它们代入

3 tan (x) + 2 sec (x) = 1

检验解是否符合

去除与方程不符的解。

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

3 tan (x) + 2 sec (x) = 1

代入

x = \frac{4 π}{3} + 2 π 1

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

整理后得

- 1 = 1

\Rightarrow 假

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

3 tan (x) + 2 sec (x) = 1

代入

x = \frac{5 π}{3} + 2 π 1

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

整理后得

1 = 1

\Rightarrow 真

x = \frac{5 π}{3} + 2 πn

作图

流行的例子

cos^2(a)=2cos^2(a)-1

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

tan(3b+14)=cot(5b+4)

tan (3 b + 14) = cot (5 b + 4)

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

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

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

sec^{2} (x) + 2 tan^{2} (x) = 2

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