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 >

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

解答

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

解答

x = \frac{π}{4} + πn

+1

度数

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

求解步骤

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

分解

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

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

因式分解出通项

- cos^{2} (x)

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

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

分别求解每个部分

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

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

cos^{2} (x) = 0

使用法则

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

cos (x) = 0

cos (x) = 0

的通解

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{π}{4} + πn

因为方程对以下值无定义:

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

x = \frac{π}{4} + πn

作图

流行的例子

sqrt(3)tan(x)-1=0,0<= x<= 2pi

3 tan (x) - 1 = 0, 0 \leq x \leq 2 π

6 + 12 sin (θ) = 0

solvefor t,cos(2t)=sin(t)

so l v e f or t, cos (2 t) = sin (t)

tan (x) = 1.8

tan (x) = 1.6