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受欢迎的三角函数 >

2sin(θ)=tan(θ)

解答

2 sin (θ) = tan (θ)

解答

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (θ) = tan (θ)

两边减去

tan (θ)

2 sin (θ) - tan (θ) = 0

用 sin, cos 表示

- tan (θ) + 2 sin (θ)

使用基本三角恒等式:

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

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

化简

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

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

将项转换为分式:

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

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

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

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

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

- sin (θ) + 2 cos (θ) sin (θ) = 0

分解

- sin (θ) + 2 cos (θ) sin (θ) : sin (θ) (2 cos (θ) - 1)

- sin (θ) + 2 cos (θ) sin (θ)

因式分解出通项

sin (θ)

= sin (θ) (- 1 + 2 cos (θ))

sin (θ) (2 cos (θ) - 1) = 0

分别求解每个部分

sin (θ) = 0 or 2 cos (θ) - 1 = 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

2 cos (θ) - 1 = 0 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

2 cos (θ) - 1 = 0

将

1

到右边

2 cos (θ) - 1 = 0

两边加上

1

2 cos (θ) - 1 + 1 = 0 + 1

化简

2 cos (θ) = 1

2 cos (θ) = 1

两边除以

2

2 cos (θ) = 1

两边除以

2

\frac{2 cos ( θ )}{2} = \frac{1}{2}

化简

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{2}

cos (θ) = \frac{1}{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}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

合并所有解

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

作图

流行的例子

sec (θ) = - 2.5

cos(4x)=0,0<= x<= 2pi

cos (4 x) = 0, 0 \leq x \leq 2 π

(csc(x))/(cot(x))=csc(x)sec(x)

\frac{csc ( x )}{cot ( x )} = csc (x) sec (x)

sin(θ)= 1/(sqrt(3))

sin (θ) = \frac{1}{3}

cos (θ) = 0.9