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受欢迎的三角函数 >

4tan^2(θ)=12,0<= θ<2pi

解答

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

解答

θ = \frac{π}{3}, θ = \frac{4 π}{3}, θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

+1

度数

θ = 6 0^{\circ}, θ = 24 0^{\circ}, θ = 12 0^{\circ}, θ = 30 0^{\circ}

求解步骤

4 tan^{2} (θ) = 12, 0 \leq θ < 2 π

用替代法求解

4 tan^{2} (θ) = 12

令

: tan (θ) = u

4 u^{2} = 12

4 u^{2} = 12 : u = 3, u = - 3

4 u^{2} = 12

两边除以

4

4 u^{2} = 12

两边除以

4

\frac{4 u ^{2}}{4} = \frac{12}{4}

化简

u^{2} = 3

u^{2} = 3

对于

x^{2} = f (a)

解为

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

u = 3, u = - 3

u = tan (θ)

代回

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3, tan (θ) = - 3

tan (θ) = 3, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{4 π}{3}

tan (θ) = 3, 0 \leq θ < 2 π

tan (θ) = 3

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}

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

在

0 \leq θ < 2 π

范围内的解

θ = \frac{π}{3}, θ = \frac{4 π}{3}

tan (θ) = - 3, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

tan (θ) = - 3, 0 \leq θ < 2 π

tan (θ) = - 3

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}

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

在

0 \leq θ < 2 π

范围内的解

θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

合并所有解

θ = \frac{π}{3}, θ = \frac{4 π}{3}, θ = \frac{2 π}{3}, θ = \frac{5 π}{3}

作图

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