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受欢迎的三角函数 >

3tan((θ+11pi)/6)=3

解答

3 tan (\frac{θ + 11 π}{6}) = 3

解答

θ = \frac{3 π}{2} + 6 πn - 11 π

+1

度数

θ = - 171 0^{\circ} + 108 0^{\circ} n

求解步骤

3 tan (\frac{θ + 11 π}{6}) = 3

两边除以

3

3 tan (\frac{θ + 11 π}{6}) = 3

两边除以

3

\frac{3 tan ( \frac{θ + 11 π}{6} )}{3} = \frac{3}{3}

化简

tan (\frac{θ + 11 π}{6}) = 1

tan (\frac{θ + 11 π}{6}) = 1

tan (\frac{θ + 11 π}{6}) = 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}

\frac{θ + 11 π}{6} = \frac{π}{4} + πn

\frac{θ + 11 π}{6} = \frac{π}{4} + πn

解

\frac{θ + 11 π}{6} = \frac{π}{4} + πn : θ = \frac{3 π}{2} + 6 πn - 11 π

\frac{θ + 11 π}{6} = \frac{π}{4} + πn

在两边乘以

6

\frac{θ + 11 π}{6} = \frac{π}{4} + πn

在两边乘以

6

\frac{6 ( θ + 11 π )}{6} = 6 \cdot \frac{π}{4} + 6 πn

化简

\frac{6 ( θ + 11 π )}{6} = 6 \cdot \frac{π}{4} + 6 πn

化简

\frac{6 ( θ + 11 π )}{6} : θ + 11 π

\frac{6 ( θ + 11 π )}{6}

数字相除:

\frac{6}{6} = 1

= θ + 11 π

化简

6 \cdot \frac{π}{4} + 6 πn : \frac{3 π}{2} + 6 πn

6 \cdot \frac{π}{4} + 6 πn

6 \cdot \frac{π}{4} = \frac{3 π}{2}

6 \cdot \frac{π}{4}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 6}{4}

约分:

2

= \frac{3 π}{2}

= \frac{3 π}{2} + 6 πn

θ + 11 π = \frac{3 π}{2} + 6 πn

θ + 11 π = \frac{3 π}{2} + 6 πn

θ + 11 π = \frac{3 π}{2} + 6 πn

将

11 π

到右边

θ + 11 π = \frac{3 π}{2} + 6 πn

两边减去

11 π

θ + 11 π - 11 π = \frac{3 π}{2} + 6 πn - 11 π

化简

θ = \frac{3 π}{2} + 6 πn - 11 π

θ = \frac{3 π}{2} + 6 πn - 11 π

θ = \frac{3 π}{2} + 6 πn - 11 π

作图

流行的例子

8cos^2(θ)+13=14cos(θ)+8

8 cos^{2} (θ) + 13 = 14 cos (θ) + 8

cos(3x)=-1/2 ,0<= x<2pi

cos (3 x) = - \frac{1}{2}, 0 \leq x < 2 π

cos (5 t) = 0

tan(4x)=(sqrt(3))/3

tan (4 x) = \frac{3}{3}

tan^2(x)=tan^2(x)sec(x)

tan^{2} (x) = tan^{2} (x) sec (x)