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受欢迎的三角函数 >

(cos^2(x)-1)/(sin^2(x))=-tan(x)

解答

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

解答

x = \frac{π}{4} + πn

+1

度数

x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

- tan (x)

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

化简

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

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

将项转换为分式:

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

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

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

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

分解

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

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

因式分解出通项

sin^{2} (x)

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

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

分别求解每个部分

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

sin^{2} (x) = 0 : x = 2 πn, x = π + 2 πn

sin^{2} (x) = 0

使用法则

x^{n} = 0 \Rightarrow x = 0

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

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

tan (x) - 1 = 0

将

1

到右边

tan (x) - 1 = 0

两边加上

1

tan (x) - 1 + 1 = 0 + 1

化简

tan (x) = 1

tan (x) = 1

tan (x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

合并所有解

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

因为方程对以下值无定义:

2 πn, π + 2 πn

x = \frac{π}{4} + πn

作图

流行的例子

cos^6(x)=-cos^2(x)

cos^{6} (x) = - cos^{2} (x)

2sin^3(x)-5sin^2(x)+2sin(x)=0

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

(cos^2(a)-3cos(a)+2)/(sin^2(a))=1

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

(sin(x)-(sqrt(2)))/2 =0

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

cos(2x)=5-6cos^2(x)

cos (2 x) = 5 - 6 cos^{2} (x)