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受欢迎的三角函数 >

cos(2θ)+6sin^2(θ)=3

解答

cos (2 θ) + 6 sin^{2} (θ) = 3

解答

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

+1

度数

θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

求解步骤

cos (2 θ) + 6 sin^{2} (θ) = 3

两边减去

3

cos (2 θ) + 6 sin^{2} (θ) - 3 = 0

使用三角恒等式改写

- 3 + cos (2 θ) + 6 sin^{2} (θ)

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

= - 3 + 1 - 2 sin^{2} (θ) + 6 sin^{2} (θ)

化简

- 3 + 1 - 2 sin^{2} (θ) + 6 sin^{2} (θ) : 4 sin^{2} (θ) - 2

- 3 + 1 - 2 sin^{2} (θ) + 6 sin^{2} (θ)

同类项相加:

- 2 sin^{2} (θ) + 6 sin^{2} (θ) = 4 sin^{2} (θ)

= - 3 + 1 + 4 sin^{2} (θ)

数字相加/相减:

- 3 + 1 = - 2

= 4 sin^{2} (θ) - 2

= 4 sin^{2} (θ) - 2

- 2 + 4 sin^{2} (θ) = 0

用替代法求解

- 2 + 4 sin^{2} (θ) = 0

令

: sin (θ) = u

- 2 + 4 u^{2} = 0

- 2 + 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

- 2 + 4 u^{2} = 0

将

2

到右边

- 2 + 4 u^{2} = 0

两边加上

2

- 2 + 4 u^{2} + 2 = 0 + 2

化简

4 u^{2} = 2

4 u^{2} = 2

两边除以

4

4 u^{2} = 2

两边除以

4

\frac{4 u ^{2}}{4} = \frac{2}{4}

化简

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

u = sin (θ)

代回

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{1}{2}, sin (θ) = - \frac{1}{2}

sin (θ) = \frac{1}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

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 = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

sin (θ) = - \frac{1}{2} : θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

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 = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

合并所有解

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

作图

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