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 >

a=((1+sin^2(x)))/((1-sin^2(x)))

解答

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

解答

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

求解步骤

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

交换两边

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

用替代法求解

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

令

: sin (x) = u

\frac{1 + u ^{2}}{1 - u ^{2}} = a

\frac{1 + u ^{2}}{1 - u ^{2}} = a : u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

\frac{1 + u ^{2}}{1 - u ^{2}} = a

在两边乘以

1 - u^{2}

\frac{1 + u ^{2}}{1 - u ^{2}} = a

在两边乘以

1 - u^{2}

\frac{1 + u ^{2}}{1 - u ^{2}} (1 - u^{2}) = a (1 - u^{2})

化简

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

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

解

1 + u^{2} = a (1 - u^{2}) : u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

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

将

1

到右边

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

两边减去

1

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

化简

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

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

将

a (1 - u^{2})

para o lado esquerdo

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

两边减去

a (1 - u^{2})

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

化简

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

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

乘开

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

- a (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

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

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

使用加减运算法则

- (- a) = a

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

乘以:

1 \cdot a = a

= - a + a u^{2}

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

将

a

到右边

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

两边加上

a

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

化简

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

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

分解

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

u^{2} + a u^{2}

因式分解出通项

u^{2}

= u^{2} (1 + a)

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

两边除以

1 + a; a \neq = - 1

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

两边除以

1 + a; a \neq = - 1

\frac{u ^{2} ( 1 + a )}{1 + a} = - \frac{1}{1 + a} + \frac{a}{1 + a}; a \neq = - 1

化简

\frac{u ^{2} ( 1 + a )}{1 + a} = - \frac{1}{1 + a} + \frac{a}{1 + a}

化简

\frac{u ^{2} ( 1 + a )}{1 + a} : u^{2}

\frac{u ^{2} ( 1 + a )}{1 + a}

约分:

1 + a

= u^{2}

化简

- \frac{1}{1 + a} + \frac{a}{1 + a} : \frac{- 1 + a}{1 + a}

- \frac{1}{1 + a} + \frac{a}{1 + a}

使用法则

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

= \frac{- 1 + a}{1 + a}

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

u^{2} = \frac{- 1 + a}{1 + a}; a \neq = - 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

u = \frac{- 1 + a}{1 + a}, u = - \frac{- 1 + a}{1 + a}; a \neq = - 1

u = sin (x)

代回

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

sin (x) = \frac{- 1 + a}{1 + a}, sin (x) = - \frac{- 1 + a}{1 + a}; a \neq = - 1

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

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn

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

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

使用反三角函数性质

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

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

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

合并所有解

x = arcsin (\frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = arcsin (- \frac{- 1 + a}{1 + a}) + 2 πn, x = π + arcsin (\frac{- 1 + a}{1 + a}) + 2 πn

作图

流行的例子

solvefor x,b*f=sin^3(x)

so l v e f or x, b \cdot f = sin^{3} (x)

2sin^2(x)+sin^3(x)-1=0

2 sin^{2} (x) + sin^{3} (x) - 1 = 0

2cos^2(x)=3cos(x)-1

2 cos^{2} (x) = 3 cos (x) - 1

sin^2(x)-4sin(x)+4=0

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

3cos^2(x)-10cos(x)+3=0

3 cos^{2} (x) - 10 cos (x) + 3 = 0