zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

tan(x)sec(x)=sqrt(2)

解答

tan (x) sec (x) = 2

解答

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

+1

度数

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

求解步骤

tan (x) sec (x) = 2

两边减去

2

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

用 sin, cos 表示

- 2 + sec (x) tan (x)

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

分式相乘:

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

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

乘以:

1 \cdot sin (x) = sin (x)

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

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

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)

= cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= cos^{2} (x)

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

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

将项转换为分式:

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

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

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

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

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

展开

u - (1 - u^{2}) 2 : u - 2 + 2 u^{2}

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

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

乘开

- 2 (1 - u^{2}) : - 2 + 2 u^{2}

- 2 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = u^{2}

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

使用加减运算法则

- (- a) = a

= - 1 \cdot 2 + 2 u^{2}

乘以:

1 \cdot 2 = 2

= - 2 + 2 u^{2}

= u - 2 + 2 u^{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

1^{2} - 42 (- 2) = 3

1^{2} - 42 (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 42 (- 2)

使用法则

- (- a) = a

= 1 + 422

422 = 8

422

使用根式运算法则:

a a = a

22 = 2

= 4 \cdot 2

数字相乘:

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

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

数字相加/相减:

- 1 + 3 = 2

= \frac{2}{2 2}

数字相除:

\frac{2}{2} = 1

= \frac{1}{2}

\frac{1}{2}

有理化:

\frac{2}{2}

\frac{1}{2}

乘以共轭根式

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

使用根式运算法则:

a a = a

22 = 2

= 2

= \frac{2}{2}

= \frac{2}{2}

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

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

数字相减:

- 1 - 3 = - 4

= \frac{- 4}{2 2}

使用分式法则:

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

= - \frac{4}{2 2}

数字相除:

\frac{4}{2} = 2

= \frac{2}{2}

使用根式运算法则:

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

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

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

使用指数法则:

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

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

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

数字相减:

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

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

使用根式运算法则:

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

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

= - 2

二次方程组的解是:

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

u = sin (x)

代回

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

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

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

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

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

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

sin (x) = - 2 :

无解

sin (x) = - 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

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

作图

流行的例子

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

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

sqrt(3)sin(x)=1+cos(x)

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

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

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

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

(sin(x))/(cos(x))=1

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