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受欢迎的三角函数 >

-9cos^2(θ)+14sin(θ)+14=0

解答

- 9 cos^{2} (θ) + 14 sin (θ) + 14 = 0

解答

θ = - 0.58903 \dots + 2 πn, θ = π + 0.58903 \dots + 2 πn, θ = \frac{3 π}{2} + 2 πn

+1

度数

θ = - 33.74898 \dots^{\circ} + 36 0^{\circ} n, θ = 213.74898 \dots^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

求解步骤

- 9 cos^{2} (θ) + 14 sin (θ) + 14 = 0

使用三角恒等式改写

14 + 14 sin (θ) - 9 cos^{2} (θ)

使用毕达哥拉斯恒等式:

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

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

= 14 + 14 sin (θ) - 9 (1 - sin^{2} (θ))

化简

14 + 14 sin (θ) - 9 (1 - sin^{2} (θ)) : 9 sin^{2} (θ) + 14 sin (θ) + 5

14 + 14 sin (θ) - 9 (1 - sin^{2} (θ))

乘开

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

- 9 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 9, b = 1, c = sin^{2} (θ)

= - 9 \cdot 1 - (- 9) sin^{2} (θ)

使用加减运算法则

- (- a) = a

= - 9 \cdot 1 + 9 sin^{2} (θ)

数字相乘:

9 \cdot 1 = 9

= - 9 + 9 sin^{2} (θ)

= 14 + 14 sin (θ) - 9 + 9 sin^{2} (θ)

化简

14 + 14 sin (θ) - 9 + 9 sin^{2} (θ) : 9 sin^{2} (θ) + 14 sin (θ) + 5

14 + 14 sin (θ) - 9 + 9 sin^{2} (θ)

对同类项分组

= 14 sin (θ) + 9 sin^{2} (θ) + 14 - 9

数字相加/相减:

14 - 9 = 5

= 9 sin^{2} (θ) + 14 sin (θ) + 5

= 9 sin^{2} (θ) + 14 sin (θ) + 5

= 9 sin^{2} (θ) + 14 sin (θ) + 5

5 + 14 sin (θ) + 9 sin^{2} (θ) = 0

用替代法求解

5 + 14 sin (θ) + 9 sin^{2} (θ) = 0

令

: sin (θ) = u

5 + 14 u + 9 u^{2} = 0

5 + 14 u + 9 u^{2} = 0 : u = - \frac{5}{9}, u = - 1

5 + 14 u + 9 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

9 u^{2} + 14 u + 5 = 0

使用求根公式求解

9 u^{2} + 14 u + 5 = 0

二次方程求根公式:

若

a = 9, b = 14, c = 5

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 9 \cdot 5}{2 \cdot 9}

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 9 \cdot 5}{2 \cdot 9}

1 4^{2} - 4 \cdot 9 \cdot 5 = 4

1 4^{2} - 4 \cdot 9 \cdot 5

数字相乘:

4 \cdot 9 \cdot 5 = 180

= 1 4^{2} - 180

1 4^{2} = 196

= 196 - 180

数字相减:

196 - 180 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- 14 \pm 4}{2 \cdot 9}

将解分隔开

u_{1} = \frac{- 14 + 4}{2 \cdot 9}, u_{2} = \frac{- 14 - 4}{2 \cdot 9}

u = \frac{- 14 + 4}{2 \cdot 9} : - \frac{5}{9}

\frac{- 14 + 4}{2 \cdot 9}

数字相加/相减:

- 14 + 4 = - 10

= \frac{- 10}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{- 10}{18}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{10}{18}

约分:

2

= - \frac{5}{9}

u = \frac{- 14 - 4}{2 \cdot 9} : - 1

\frac{- 14 - 4}{2 \cdot 9}

数字相减:

- 14 - 4 = - 18

= \frac{- 18}{2 \cdot 9}

数字相乘:

2 \cdot 9 = 18

= \frac{- 18}{18}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{18}{18}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

u = - \frac{5}{9}, u = - 1

u = sin (θ)

代回

sin (θ) = - \frac{5}{9}, sin (θ) = - 1

sin (θ) = - \frac{5}{9}, sin (θ) = - 1

sin (θ) = - \frac{5}{9} : θ = arcsin (- \frac{5}{9}) + 2 πn, θ = π + arcsin (\frac{5}{9}) + 2 πn

sin (θ) = - \frac{5}{9}

使用反三角函数性质

sin (θ) = - \frac{5}{9}

sin (θ) = - \frac{5}{9}

的通解

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{5}{9}) + 2 πn, θ = π + arcsin (\frac{5}{9}) + 2 πn

θ = arcsin (- \frac{5}{9}) + 2 πn, θ = π + arcsin (\frac{5}{9}) + 2 πn

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

sin (θ) = - 1

sin (θ) = - 1

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{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

合并所有解

θ = arcsin (- \frac{5}{9}) + 2 πn, θ = π + arcsin (\frac{5}{9}) + 2 πn, θ = \frac{3 π}{2} + 2 πn

以小数形式表示解

θ = - 0.58903 \dots + 2 πn, θ = π + 0.58903 \dots + 2 πn, θ = \frac{3 π}{2} + 2 πn

作图

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