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受欢迎的三角函数 >

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

解答

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

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

使用三角恒等式改写

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

使用基本三角恒等式:

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

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

使用指数法则:

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

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

化简

- 1 + 1 - sin^{2} (x) + \frac{sin ^{2} ( x )}{1 - sin ^{2} ( x )} : \frac{sin ^{4} ( x )}{- sin ^{2} ( x ) + 1}

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

- 1 + 1 = 0

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

将项转换为分式:

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

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

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

乘开

sin^{2} (x) - sin^{2} (x) (- sin^{2} (x) + 1) : sin^{4} (x)

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

乘开

- sin^{2} (x) (- sin^{2} (x) + 1) : sin^{4} (x) - sin^{2} (x)

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

使用分配律:

a (b + c) = ab + a c

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

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

使用加减运算法则

- (- a) = a, + (- a) = - a

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

化简

sin^{2} (x) sin^{2} (x) - 1 \cdot sin^{2} (x) : sin^{4} (x) - sin^{2} (x)

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

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

1 \cdot sin^{2} (x)

乘以:

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

= sin^{2} (x)

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

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

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

同类项相加:

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

= sin^{4} (x)

= \frac{sin ^{4} ( x )}{- sin ^{2} ( x ) + 1}

= \frac{sin ^{4} ( x )}{- sin ^{2} ( x ) + 1}

\frac{sin ^{4} ( x )}{1 - sin ^{2} ( x )} = 0

用替代法求解

\frac{sin ^{4} ( x )}{1 - sin ^{2} ( x )} = 0

令

: sin (x) = u

\frac{u ^{4}}{1 - u ^{2}} = 0

\frac{u ^{4}}{1 - u ^{2}} = 0 : u = 0

\frac{u ^{4}}{1 - u ^{2}} = 0

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

u^{4} = 0

解

u^{4} = 0 : u = 0

u^{4} = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0

解是

u = 0

u = 0

验证解

找到无定义的点（奇点）:

u = 1, u = - 1

取

\frac{u ^{4}}{1 - u ^{2}}

的分母，令其等于零

解

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

1 - u^{2} = 0

将

1

到右边

1 - u^{2} = 0

两边减去

1

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

化简

- u^{2} = - 1

- u^{2} = - 1

两边除以

- 1

- u^{2} = - 1

两边除以

- 1

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

化简

u^{2} = 1

u^{2} = 1

对于

x^{2} = f (a)

解为

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

u = 1, u = - 1

1 = 1

1

使用根式运算法则:

1 = 1

= 1

- 1 = - 1

- 1

使用根式运算法则:

1 = 1

1 = 1

= - 1

u = 1, u = - 1

以下点无定义

u = 1, u = - 1

将不在定义域的点与解相综合:

u = 0

u = sin (x)

代回

sin (x) = 0

sin (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

合并所有解

x = 2 πn, x = π + 2 πn

作图

流行的例子

cot^3(x)+cot(x)=0

21+18cos(x)=16(1-cos^2(x))

sin(2a+10)=cos(3a-20)

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4sin^2(x)+cos(x)+1=0