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受欢迎的三角函数 >

tan^2(x)-1=0

解答

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

解答

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

+1

度数

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

求解步骤

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

用替代法求解

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

令

: tan (x) = u

u^{2} - 1 = 0

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

u^{2} - 1 = 0

将

1

到右边

u^{2} - 1 = 0

两边加上

1

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

使用法则

1 = 1

= 1

- 1 = - 1

- 1

使用法则

1 = 1

= - 1

u = 1, u = - 1

u = tan (x)

代回

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

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

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

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

合并所有解

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

作图

流行的例子

cos(2θ)=-(sqrt(3))/2

cot^2(135)-sin(210)+5cos^2(225)

2sin^2(θ)=1-sin(θ)

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