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受欢迎的三角函数 >

2sin(θ)=sqrt(3)tan(θ)

解答

2 sin (θ) = 3 tan (θ)

解答

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

+1

度数

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

求解步骤

2 sin (θ) = 3 tan (θ)

两边减去

3 tan (θ)

2 sin (θ) - 3 tan (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 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

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

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

- 3 + 2 cos (θ) = 0

将

3

到右边

- 3 + 2 cos (θ) = 0

两边加上

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

合并所有解

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

作图

流行的例子

sin (θ) = - 0.78

4tan(θ)=5-tan(θ)

4 tan (θ) = 5 - tan (θ)

tan (θ) = \frac{15}{6}

-cos(2x)=cos^2(x)-sin^2(x)

- cos (2 x) = cos^{2} (x) - sin^{2} (x)

tan (θ) = \frac{15}{9}