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受欢迎的三角函数 >

tan^3(x)-1tan(x)=0

解答

tan^{3} (x) - 1 tan (x) = 0

解答

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

+1

度数

x = 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

求解步骤

tan^{3} (x) - 1 \cdot tan (x) = 0

用替代法求解

tan^{3} (x) - 1 \cdot tan (x) = 0

令

: tan (x) = u

u^{3} - 1 \cdot u = 0

u^{3} - 1 \cdot u = 0 : u = 0, u = - 1, u = 1

u^{3} - 1 \cdot u = 0

因式分解

u^{3} - 1 \cdot u : u (u + 1) (u - 1)

u^{3} - 1 \cdot u

因式分解出通项

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

u^{3} - u

使用指数法则:

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

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

= u^{2} u - u

因式分解出通项

u

= u (u^{2} - 1)

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

u (u + 1) (u - 1) = 0

使用零因数法则： If

ab = 0

then

a = 0

or

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

解

u + 1 = 0 : u = - 1

u + 1 = 0

将

1

到右边

u + 1 = 0

两边减去

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

两边加上

1

u - 1 + 1 = 0 + 1

化简

u = 1

u = 1

解为

u = 0, u = - 1, u = 1

u = tan (x)

代回

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

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

tan (x) = 0 : x = πn

tan (x) = 0

tan (x) = 0

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 = 0 + πn

x = 0 + πn

解

x = 0 + πn : x = πn

x = 0 + πn

0 + πn = πn

x = πn

x = πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

tan (x) = 1 : x = \frac{π}{4} + πn

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

作图

流行的例子

2cos(x)-1=0,0<= x<= 2pi

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

tan (2 x) = 2

sin(2x)=cos(2x)+1

sin (2 x) = cos (2 x) + 1

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

solvefor y,sin(x)=x(1+tan(y))

so l v e f or y, sin (x) = x (1 + tan (y))