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受欢迎的三角函数 >

-3cos^2(u)sin(u)-2sin(2u)=0

解答

- 3 cos^{2} (u) sin (u) - 2 sin (2 u) = 0

解答

u = 2 πn, u = π + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

+1

度数

u = 0^{\circ} + 36 0^{\circ} n, u = 18 0^{\circ} + 36 0^{\circ} n, u = 9 0^{\circ} + 36 0^{\circ} n, u = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

- 3 cos^{2} (u) sin (u) - 2 sin (2 u) = 0

使用三角恒等式改写

- 2 sin (2 u) - 3 cos^{2} (u) sin (u)

使用倍角公式:

sin (2 x) = 2 sin (x) cos (x)

= - 2 \cdot 2 sin (u) cos (u) - 3 cos^{2} (u) sin (u)

化简

= - 4 sin (u) cos (u) - 3 cos^{2} (u) sin (u)

- 3 cos^{2} (u) sin (u) - 4 cos (u) sin (u) = 0

分解

- 3 cos^{2} (u) sin (u) - 4 cos (u) sin (u) : - sin (u) cos (u) (3 cos (u) + 4)

- 3 cos^{2} (u) sin (u) - 4 cos (u) sin (u)

使用指数法则:

a^{b + c} = a^{b} a^{c}

sin (u) cos^{2} (u) = sin (u) cos (u) cos (u)

= - 3 sin (u) cos (u) cos (u) - 4 sin (u) cos (u)

因式分解出通项

- sin (u) cos (u)

= - sin (u) cos (u) (3 cos (u) + 4)

- sin (u) cos (u) (3 cos (u) + 4) = 0

分别求解每个部分

sin (u) = 0 or cos (u) = 0 or 3 cos (u) + 4 = 0

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = 0

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

解

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

cos (u) = 0

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}

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

3 cos (u) + 4 = 0 :

无解

3 cos (u) + 4 = 0

将

4

到右边

3 cos (u) + 4 = 0

两边减去

4

3 cos (u) + 4 - 4 = 0 - 4

化简

3 cos (u) = - 4

3 cos (u) = - 4

两边除以

3

3 cos (u) = - 4

两边除以

3

\frac{3 cos ( u )}{3} = \frac{- 4}{3}

化简

cos (u) = - \frac{4}{3}

cos (u) = - \frac{4}{3}

- 1 \leq cos (x) \leq 1

无解

合并所有解

u = 2 πn, u = π + 2 πn, u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

作图

流行的例子

cot (θ) = - \frac{7}{24}

2sin(x)+2cos(x)=sqrt(6)

2 sin (x) + 2 cos (x) = 6

0=-0.65sin(13x)

0 = - 0.65 sin (13 x)

solvefor x,tan^2(x)-1=0

so l v e f or x, tan^{2} (x) - 1 = 0

c^9=3tan(-3x+c)

c^{9} = 3 tan (- 3 x + c)