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受欢迎的三角函数 >

1/(6tan^6(x))= 1/(6sec^6(x))

解答

\frac{1}{6 tan ^{6} ( x )} = \frac{1}{6 sec ^{6} ( x )}

解答

x \in R 无解

求解步骤

\frac{1}{6 tan ^{6} ( x )} = \frac{1}{6 sec ^{6} ( x )}

两边减去

\frac{1}{6 sec ^{6} ( x )}

\frac{1}{6 tan ^{6} ( x )} - \frac{1}{6 sec ^{6} ( x )} = 0

化简

\frac{1}{6 tan ^{6} ( x )} - \frac{1}{6 sec ^{6} ( x )} : \frac{sec ^{6} ( x ) - tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

\frac{1}{6 tan ^{6} ( x )} - \frac{1}{6 sec ^{6} ( x )}

6 tan^{6} (x), 6 sec^{6} (x)

的最小公倍数:

6 tan^{6} (x) sec^{6} (x)

6 tan^{6} (x), 6 sec^{6} (x)

最小公倍数 (LCM)

6, 6

的最小公倍数:

6

6, 6

最小公倍数 (LCM)

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

6

质因数分解:

2 \cdot 3

6

6

除以

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 3

将每个因子乘以它在

6

或

6

中出现的最多次数

= 2 \cdot 3

数字相乘:

2 \cdot 3 = 6

= 6

计算出由出现在

6 tan^{6} (x)

或

6 sec^{6} (x)

中的因子组成的表达式

= 6 tan^{6} (x) sec^{6} (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

6 tan^{6} (x) sec^{6} (x)

对于

\frac{1}{6 tan ^{6} ( x )} :

将分母和分子乘以

sec^{6} (x)

\frac{1}{6 tan ^{6} ( x )} = \frac{1 \cdot sec ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )} = \frac{sec ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

对于

\frac{1}{6 sec ^{6} ( x )} :

将分母和分子乘以

tan^{6} (x)

\frac{1}{6 sec ^{6} ( x )} = \frac{1 \cdot tan ^{6} ( x )}{6 sec ^{6} ( x ) tan ^{6} ( x )} = \frac{tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

= \frac{sec ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )} - \frac{tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( 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{sec ^{6} ( x ) - tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )}

\frac{sec ^{6} ( x ) - tan ^{6} ( x )}{6 tan ^{6} ( x ) sec ^{6} ( x )} = 0

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

sec^{6} (x) - tan^{6} (x) = 0

分解

sec^{6} (x) - tan^{6} (x) : (sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x)) (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x))

sec^{6} (x) - tan^{6} (x)

将

sec^{6} (x) - tan^{6} (x)

改写为

(sec^{3} (x))^{2} - (tan^{3} (x))^{2}

sec^{6} (x) - tan^{6} (x)

使用指数法则:

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

sec^{6} (x) = (sec^{3} (x))^{2}

= (sec^{3} (x))^{2} - tan^{6} (x)

使用指数法则:

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

tan^{6} (x) = (tan^{3} (x))^{2}

= (sec^{3} (x))^{2} - (tan^{3} (x))^{2}

= (sec^{3} (x))^{2} - (tan^{3} (x))^{2}

使用平方差公式:

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

(sec^{3} (x))^{2} - (tan^{3} (x))^{2} = (sec^{3} (x) + tan^{3} (x)) (sec^{3} (x) - tan^{3} (x))

= (sec^{3} (x) + tan^{3} (x)) (sec^{3} (x) - tan^{3} (x))

分解

sec^{3} (x) + tan^{3} (x) : (sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x))

sec^{3} (x) + tan^{3} (x)

使用立方和公式:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

sec^{3} (x) + tan^{3} (x) = (sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x))

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

= (sec (x) + tan (x)) (sec^{2} (x) - sec (x) tan (x) + tan^{2} (x)) (sec^{3} (x) - tan^{3} (x))

分解

sec^{3} (x) - tan^{3} (x) : (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x))

sec^{3} (x) - tan^{3} (x)

使用立方差公式:

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

sec^{3} (x) - tan^{3} (x) = (sec (x) - tan (x)) (sec^{2} (x) + sec (x) tan (x) + tan^{2} (x))

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

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

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

使用三角恒等式改写

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

(sec (x) + tan (x)) (sec (x) - tan (x)) = 1

(sec (x) + tan (x)) (sec (x) - tan (x))

乘开

(sec (x) + tan (x)) (sec (x) - tan (x)) : sec^{2} (x) - tan^{2} (x)

(sec (x) + tan (x)) (sec (x) - tan (x))

使用平方差公式:

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

a = sec (x), b = tan (x)

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

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

使用毕达哥拉斯恒等式:

sec^{2} (x) = tan^{2} (x) + 1

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

= 1

= 1 \cdot (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x))

化简

1 \cdot (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x)) : (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x))

1 \cdot (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) (sec^{2} (x) + tan^{2} (x) - sec (x) tan (x))

乘以:

1 \cdot (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x)) = (sec^{2} (x) + tan^{2} (x) + sec (x) tan (x))

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

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

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

分别求解每个部分

sec^{2} (x) + tan^{2} (x) + sec (x) tan (x) = 0 or sec^{2} (x) + tan^{2} (x) - sec (x) tan (x) = 0

sec^{2} (x) + tan^{2} (x) + sec (x) tan (x) = 0 :

无解

sec^{2} (x) + tan^{2} (x) + sec (x) tan (x) = 0

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

使用指数法则:

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

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

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

使用法则

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

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

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

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

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

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

用替代法求解

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

令

: sin (x) = u

1 + u + u^{2} = 0

1 + u + u^{2} = 0 : u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

1 + u + u^{2} = 0

改写成标准形式

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

u^{2} + u + 1 = 0

使用求根公式求解

u^{2} + u + 1 = 0

二次方程求根公式:

若

a = 1, b = 1, c = 1

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

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

化简

1^{2} - 4 \cdot 1 \cdot 1 : 3 i

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

使用根式运算法则:

- a = - 1 a

- 3 = - 1 3

= - 1 3

使用虚数运算法则:

- 1 = i

= 3 i

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

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

将

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

改写成标准复数形式:

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

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

使用分式法则:

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

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

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

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

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

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

数字相乘:

2 \cdot 1 = 2

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

将

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

改写成标准复数形式:

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

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

使用分式法则:

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

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

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

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

二次方程组的解是:

u = - \frac{1}{2} + i \frac{3}{2}, u = - \frac{1}{2} - i \frac{3}{2}

u = sin (x)

代回

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

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

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

无解

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

无解

sin (x) = - \frac{1}{2} - i \frac{3}{2} :

无解

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

无解

合并所有解

无解

sec^{2} (x) + tan^{2} (x) - sec (x) tan (x) = 0 :

无解

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

使用基本三角恒等式:

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

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

化简

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

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

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

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

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

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

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

使用指数法则:

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

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

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

使用法则

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

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

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

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

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

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

用替代法求解

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

令

: sin (x) = u

1 - u + u^{2} = 0

1 - u + u^{2} = 0 : u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

1 - u + u^{2} = 0

改写成标准形式

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

u^{2} - u + 1 = 0

使用求根公式求解

u^{2} - u + 1 = 0

二次方程求根公式:

若

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

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

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

化简

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

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

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 - 4

数字相减:

1 - 4 = - 3

= - 3

使用根式运算法则:

- a = - 1 a

- 3 = - 1 3

= - 1 3

使用虚数运算法则:

- 1 = i

= 3 i

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

将

\frac{1 + 3 i}{2}

改写成标准复数形式:

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

\frac{1 + 3 i}{2}

使用分式法则:

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

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

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

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

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

将

\frac{1 - 3 i}{2}

改写成标准复数形式:

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

\frac{1 - 3 i}{2}

使用分式法则:

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

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

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

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

二次方程组的解是:

u = \frac{1}{2} + i \frac{3}{2}, u = \frac{1}{2} - i \frac{3}{2}

u = sin (x)

代回

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

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

sin (x) = \frac{1}{2} + i \frac{3}{2} :

无解

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

无解

sin (x) = \frac{1}{2} - i \frac{3}{2} :

无解

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

无解

合并所有解

无解

合并所有解

x \in R 无解

作图

流行的例子

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

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

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

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

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

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

solvefor x,y=3cos(fxx+pi/2)+5

so l v e f or x, y = 3 cos (f xx + \frac{π}{2}) + 5

sin(x)cos(x)=sin(x),0<x<= 2pi

sin (x) cos (x) = sin (x), 0 < x \leq 2 π