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受欢迎的三角函数 >

cos(2θ)+14sin^2(θ)=10

解答

cos (2 θ) + 14 sin^{2} (θ) = 10

解答

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

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (2 θ) + 14 sin^{2} (θ) = 10

两边减去

10

cos (2 θ) + 14 sin^{2} (θ) - 10 = 0

使用三角恒等式改写

- 10 + cos (2 θ) + 14 sin^{2} (θ)

使用倍角公式:

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

= - 10 + 1 - 2 sin^{2} (θ) + 14 sin^{2} (θ)

化简

- 10 + 1 - 2 sin^{2} (θ) + 14 sin^{2} (θ) : 12 sin^{2} (θ) - 9

- 10 + 1 - 2 sin^{2} (θ) + 14 sin^{2} (θ)

同类项相加:

- 2 sin^{2} (θ) + 14 sin^{2} (θ) = 12 sin^{2} (θ)

= - 10 + 1 + 12 sin^{2} (θ)

数字相加/相减:

- 10 + 1 = - 9

= 12 sin^{2} (θ) - 9

= 12 sin^{2} (θ) - 9

- 9 + 12 sin^{2} (θ) = 0

用替代法求解

- 9 + 12 sin^{2} (θ) = 0

令

: sin (θ) = u

- 9 + 12 u^{2} = 0

- 9 + 12 u^{2} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

- 9 + 12 u^{2} = 0

将

9

到右边

- 9 + 12 u^{2} = 0

两边加上

9

- 9 + 12 u^{2} + 9 = 0 + 9

化简

12 u^{2} = 9

12 u^{2} = 9

两边除以

12

12 u^{2} = 9

两边除以

12

\frac{12 u ^{2}}{12} = \frac{9}{12}

化简

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

对于

x^{2} = f (a)

解为

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

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

化简

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

因式分解数字:

4 = 2^{2}

= 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

u = sin (θ)

代回

sin (θ) = \frac{3}{2}, sin (θ) = - \frac{3}{2}

sin (θ) = \frac{3}{2}, sin (θ) = - \frac{3}{2}

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

sin (θ) = \frac{3}{2}

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

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

sin (θ) = - \frac{3}{2} : θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sin (θ) = - \frac{3}{2}

sin (θ) = - \frac{3}{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{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

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

合并所有解

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

作图

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