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受欢迎的三角函数 >

4sin(θ)cos^2(θ)=3sin(θ)

解答

4 sin (θ) cos^{2} (θ) = 3 sin (θ)

解答

θ = 2 πn, θ = π + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 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 = 15 0^{\circ} + 36 0^{\circ} n, θ = 21 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

4 sin (θ) cos^{2} (θ) = 3 sin (θ)

两边减去

3 sin (θ)

4 cos^{2} (θ) sin (θ) - 3 sin (θ) = 0

分解

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

4 cos^{2} (θ) sin (θ) - 3 sin (θ)

因式分解出通项

sin (θ)

= sin (θ) (4 cos^{2} (θ) - 3)

分解

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

4 cos^{2} (θ) - 3

将

4 cos^{2} (θ) - 3

改写为

(2 cos (θ))^{2} - (3)^{2}

4 cos^{2} (θ) - 3

将

4

改写为

2^{2}

= 2^{2} cos^{2} (θ) - 3

使用根式运算法则:

a = (a)^{2}

3 = (3)^{2}

= 2^{2} cos^{2} (θ) - (3)^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

2^{2} cos^{2} (θ) = (2 cos (θ))^{2}

= (2 cos (θ))^{2} - (3)^{2}

= (2 cos (θ))^{2} - (3)^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(2 cos (θ))^{2} - (3)^{2} = (2 cos (θ) + 3) (2 cos (θ) - 3)

= (2 cos (θ) + 3) (2 cos (θ) - 3)

= sin (θ) (2 cos (θ) + 3) (2 cos (θ) - 3)

sin (θ) (2 cos (θ) + 3) (2 cos (θ) - 3) = 0

分别求解每个部分

sin (θ) = 0 or 2 cos (θ) + 3 = 0 or 2 cos (θ) - 3 = 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) + 3 = 0 : θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

2 cos (θ) + 3 = 0

将

3

到右边

2 cos (θ) + 3 = 0

两边减去

3

2 cos (θ) + 3 - 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{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

θ = \frac{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn

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

2 cos (θ) - 3 = 0

将

3

到右边

2 cos (θ) - 3 = 0

两边加上

3

2 cos (θ) - 3 + 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{5 π}{6} + 2 πn, θ = \frac{7 π}{6} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

作图

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