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受欢迎的三角函数 >

tan(θ)=(2sqrt(3))/3 sin(θ),0<= θ<= 2pi

解答

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ \leq 2 π

解答

θ = 0, θ = π, θ = 2 π, θ = \frac{π}{6}, θ = \frac{11 π}{6}

+1

度数

θ = 0^{\circ}, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 3 0^{\circ}, θ = 33 0^{\circ}

求解步骤

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ \leq 2 π

两边减去

\frac{2 3}{3} sin (θ)

tan (θ) - \frac{2 sin ( θ )}{3} = 0

化简

tan (θ) - \frac{2 sin ( θ )}{3} : \frac{3 tan ( θ ) - 2 sin ( θ )}{3}

tan (θ) - \frac{2 sin ( θ )}{3}

将项转换为分式:

tan (θ) = \frac{tan ( θ ) 3}{3}

= \frac{tan ( θ ) 3}{3} - \frac{2 sin ( θ )}{3}

因为分母相等，所以合并分式:

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

= \frac{tan ( θ ) 3 - 2 sin ( θ )}{3}

\frac{3 tan ( θ ) - 2 sin ( θ )}{3} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

3 tan (θ) - 2 sin (θ) = 0

用 sin, cos 表示

- 2 sin (θ) + 3 tan (θ)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= - 2 sin (θ) + 3 \frac{sin ( θ )}{cos ( θ )}

化简

- 2 sin (θ) + 3 \frac{sin ( θ )}{cos ( θ )} : \frac{- 2 sin ( θ ) cos ( θ ) + 3 sin ( θ )}{cos ( θ )}

- 2 sin (θ) + 3 \frac{sin ( θ )}{cos ( θ )}

乘

3 \frac{sin ( θ )}{cos ( θ )} : \frac{3 sin ( θ )}{cos ( θ )}

3 \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

= \frac{sin ( θ ) 3}{cos ( θ )}

= - 2 sin (θ) + \frac{3 sin ( θ )}{cos ( θ )}

将项转换为分式:

2 sin (θ) = \frac{2 sin ( θ ) cos ( θ )}{cos ( θ )}

= - \frac{2 sin ( θ ) cos ( θ )}{cos ( θ )} + \frac{sin ( θ ) 3}{cos ( θ )}

因为分母相等，所以合并分式:

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

= \frac{- 2 sin ( θ ) cos ( θ ) + sin ( θ ) 3}{cos ( θ )}

= \frac{- 2 sin ( θ ) cos ( θ ) + 3 sin ( θ )}{cos ( θ )}

\frac{sin ( θ ) 3 - 2 cos ( θ ) sin ( θ )}{cos ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (θ) 3 - 2 cos (θ) sin (θ) = 0

分解

sin (θ) 3 - 2 cos (θ) sin (θ) : sin (θ) (3 - 2 cos (θ))

sin (θ) 3 - 2 cos (θ) sin (θ)

因式分解出通项

sin (θ)

= sin (θ) (3 - 2 cos (θ))

sin (θ) (3 - 2 cos (θ)) = 0

分别求解每个部分

sin (θ) = 0 or 3 - 2 cos (θ) = 0

sin (θ) = 0, 0 \leq θ \leq 2 π : θ = 0, θ = π, θ = 2 π

sin (θ) = 0, 0 \leq θ \leq 2 π

sin (θ) = 0

的通解

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

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

在

0 \leq θ \leq 2 π

范围内的解

θ = 0, θ = π, θ = 2 π

3 - 2 cos (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{11 π}{6}

3 - 2 cos (θ) = 0, 0 \leq θ \leq 2 π

将

3

到右边

3 - 2 cos (θ) = 0

两边减去

3

3 - 2 cos (θ) - 3 = 0 - 3

化简

- 2 cos (θ) = - 3

- 2 cos (θ) = - 3

两边除以

- 2

- 2 cos (θ) = - 3

两边除以

- 2

\frac{- 2 cos ( θ )}{- 2} = \frac{- 3}{- 2}

化简

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

cos (θ) = \frac{3}{2}

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}

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

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

在

0 \leq θ \leq 2 π

范围内的解

θ = \frac{π}{6}, θ = \frac{11 π}{6}

合并所有解

θ = 0, θ = π, θ = 2 π, θ = \frac{π}{6}, θ = \frac{11 π}{6}

作图

流行的例子

solvefor x,cos(x+y)=-cos(x)

so l v e f or x, cos (x + y) = - cos (x)

4+2tan^2(x)=5tan^2(x)+3

4 + 2 tan^{2} (x) = 5 tan^{2} (x) + 3

solvefor θ,sec(θ)=2

so l v e f or θ, sec (θ) = 2

2 cos (4 θ) = 0

1032/3145 sin(4x)+2196/3145 cos(4x)=0

\frac{1032}{3145} sin (4 x) + \frac{2196}{3145} cos (4 x) = 0