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受欢迎的三角函数 >

solvefor x,cot(-x)cos(-x)+sin(-x)=(-1)/f

解答

求解

x, cot (- x) cos (- x) + sin (- x) = \frac{- 1}{f}

解答

x = arcsin (f) + 2 πn, x = π + arcsin (- f) + 2 πn

求解步骤

cot (- x) cos (- x) + sin (- x) = \frac{- 1}{f}

使用三角恒等式改写

cot (- x) cos (- x) + sin (- x) = \frac{- 1}{f}

- sin (x) + cos (x) (- cot (x)) = \frac{- 1}{f}

化简

- sin (x) + cos (x) (- cot (x)) : - sin (x) - cos (x) cot (x)

- sin (x) + cos (x) (- cot (x))

去除括号:

(- a) = - a

= - sin (x) - cos (x) cot (x)

化简

\frac{- 1}{f} : - \frac{1}{f}

\frac{- 1}{f}

使用分式法则:

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

= - \frac{1}{f}

- sin (x) - cos (x) cot (x) = - \frac{1}{f}

- sin (x) - cos (x) cot (x) = - \frac{1}{f}

两边减去

- \frac{1}{f}

- sin (x) - cos (x) cot (x) + \frac{1}{f} = 0

化简

- sin (x) - cos (x) cot (x) + \frac{1}{f} : \frac{- f sin ( x ) - f cos ( x ) cot ( x ) + 1}{f}

- sin (x) - cos (x) cot (x) + \frac{1}{f}

将项转换为分式:

sin (x) = \frac{sin ( x ) f}{f}, cos (x) cot (x) = \frac{cos ( x ) cot ( x ) f}{f}

= - \frac{sin ( x ) f}{f} - \frac{cos ( x ) cot ( x ) f}{f} + \frac{1}{f}

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 ) f - cos ( x ) cot ( x ) f + 1}{f}

\frac{- f sin ( x ) - f cos ( x ) cot ( x ) + 1}{f} = 0

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

- f sin (x) - f cos (x) cot (x) + 1 = 0

使用三角恒等式改写

1 - sin (x) f - cos (x) cot (x) f

使用基本三角恒等式:

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

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

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

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

分式相乘:

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

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

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

cos (x) cos (x) f

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) f

数字相加:

1 + 1 = 2

= cos^{2} (x) f

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

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

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

合并分式

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

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

将项转换为分式:

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

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

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

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

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

sin (x) f sin (x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= f sin^{2} (x)

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

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

乘开

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

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = - f, b = 1, c = sin^{2} (x)

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

使用加减运算法则

- (- a) = a

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

乘以:

1 \cdot f = f

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

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

同类项相加:

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

= - f

= \frac{- f}{sin ( x )}

使用分式法则:

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

= - \frac{f}{sin ( x )}

1 - \frac{f}{sin ( x )} = 0

1 - \frac{f}{sin ( x )} = 0

在两边乘以

sin (x)

1 - \frac{f}{sin ( x )} = 0

在两边乘以

sin (x)

1 \cdot sin (x) - \frac{f}{sin ( x )} sin (x) = 0 \cdot sin (x)

化简

1 \cdot sin (x) - \frac{f}{sin ( x )} sin (x) = 0 \cdot sin (x)

化简

1 \cdot sin (x) : sin (x)

1 \cdot sin (x)

乘以:

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

= sin (x)

化简

- \frac{f}{sin ( x )} sin (x) : - f

- \frac{f}{sin ( x )} sin (x)

分式相乘:

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

= - \frac{f sin ( x )}{sin ( x )}

约分:

sin (x)

= - f

化简

0 \cdot sin (x) : 0

0 \cdot sin (x)

使用法则

0 \cdot a = 0

= 0

sin (x) - f = 0

sin (x) - f = 0

sin (x) - f = 0

将

f

到右边

sin (x) - f = 0

两边加上

f

sin (x) - f + f = 0 + f

化简

sin (x) = f

sin (x) = f

使用反三角函数性质

sin (x) = f

sin (x) = f

的通解

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

x = arcsin (f) + 2 πn, x = π + arcsin (- f) + 2 πn

x = arcsin (f) + 2 πn, x = π + arcsin (- f) + 2 πn

作图

流行的例子

tan^2(x)sin(x)=3sin(x)

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

cos(θ)-sin(2θ)=0

cos (θ) - sin (2 θ) = 0

sin (7 x) = 0

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

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

solvefor x,tan(x)=-1

so l v e f or x, tan (x) = - 1