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受欢迎的三角函数 >

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

解答

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

解答

x = 2 πn, x = π + 2 πn

+1

度数

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

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

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

用 sin, cos 表示

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

使用基本三角恒等式:

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

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

化简

- sin (x) + (\frac{sin ( x )}{cos ( x )})^{2} - sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2} : \frac{- cos ^{2} ( x ) sin ( x ) + sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

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

(\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 )}

sin^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2} = \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

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

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

分式相乘:

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

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

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

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

使用指数法则:

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

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

= sin^{2 + 2} (x)

数字相加:

2 + 2 = 4

= sin^{4} (x)

= \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

= - sin (x) + \frac{sin ^{2} ( x )}{cos ^{2} ( x )} - \frac{sin ^{4} ( x )}{cos ^{2} ( x )}

合并分式

\frac{sin ^{2} ( x )}{cos ^{2} ( x )} - \frac{sin ^{4} ( x )}{cos ^{2} ( x )} : \frac{sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

使用法则

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

= \frac{sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

= - sin (x) + \frac{sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

将项转换为分式:

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

= - \frac{sin ( x ) cos ^{2} ( x )}{cos ^{2} ( x )} + \frac{sin ^{2} ( x ) - sin ^{4} ( 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{- sin ( x ) cos ^{2} ( x ) + sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

= \frac{- cos ^{2} ( x ) sin ( x ) + sin ^{2} ( x ) - sin ^{4} ( x )}{cos ^{2} ( x )}

\frac{sin ^{2} ( x ) - sin ^{4} ( x ) - cos ^{2} ( x ) sin ( x )}{cos ^{2} ( x )} = 0

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

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

分解

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x) : sin (x) (sin (x) - sin^{3} (x) - cos^{2} (x))

sin^{2} (x) - sin^{4} (x) - cos^{2} (x) sin (x)

使用指数法则:

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

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

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

因式分解出通项

sin (x)

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

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

分别求解每个部分

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) - sin^{3} (x) - cos^{2} (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

使用三角恒等式改写

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

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

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

用替代法求解

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

令

: sin (x) = u

- 1 + u + u^{2} - u^{3} = 0

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

- 1 + u + u^{2} - u^{3} = 0

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

- u^{3} + u^{2} + u - 1 = 0

因式分解

- u^{3} + u^{2} + u - 1 : - (u - 1)^{2} (u + 1)

- u^{3} + u^{2} + u - 1

因式分解出通项

- 1

= - (u^{3} - u^{2} - u + 1)

分解

u^{3} - u^{2} - u + 1 : (u - 1) (u + 1) (u - 1)

u^{3} - u^{2} - u + 1

= (u^{3} - u^{2}) + (- u + 1)

从

- u + 1

分解出因式

- 1

:

- (u - 1)

- u + 1

因式分解出通项

- 1

= - (u - 1)

从

u^{3} - u^{2}

分解出因式

u^{2}

:

u^{2} (u - 1)

u^{3} - u^{2}

使用指数法则:

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

u^{3} = u u^{2}

= u u^{2} - u^{2}

因式分解出通项

u^{2}

= u^{2} (u - 1)

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

因式分解出通项

u - 1

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

分解

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

u^{2} - 1

将

1

改写为

1^{2}

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

使用平方差公式:

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

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

= (u + 1) (u - 1)

= (u - 1) (u + 1) (u - 1)

= - (u - 1) (u + 1) (u - 1)

整理后得

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

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

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u - 1 = 0 or u + 1 = 0

解

u - 1 = 0 : u = 1

u - 1 = 0

将

1

到右边

u - 1 = 0

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

1

u + 1 - 1 = 0 - 1

化简

u = - 1

u = - 1

解为

u = 1, u = - 1

u = sin (x)

代回

sin (x) = 1, sin (x) = - 1

sin (x) = 1, sin (x) = - 1

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

sin (x) = - 1

sin (x) = - 1

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

合并所有解

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

合并所有解

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

因为方程对以下值无定义:

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn

x = 2 πn, x = π + 2 πn

作图

流行的例子

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