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受欢迎的三角函数 >

6sin^2(θ)-17sin(θ)+14=-4sin(θ)+9

解答

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

解答

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

+1

度数

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

用替代法求解

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

令

: sin (θ) = u

6 u^{2} - 17 u + 14 = - 4 u + 9

6 u^{2} - 17 u + 14 = - 4 u + 9 : u = \frac{5}{3}, u = \frac{1}{2}

6 u^{2} - 17 u + 14 = - 4 u + 9

将

9

para o lado esquerdo

6 u^{2} - 17 u + 14 = - 4 u + 9

两边减去

9

6 u^{2} - 17 u + 14 - 9 = - 4 u + 9 - 9

化简

6 u^{2} - 17 u + 5 = - 4 u

6 u^{2} - 17 u + 5 = - 4 u

将

4 u

para o lado esquerdo

6 u^{2} - 17 u + 5 = - 4 u

两边加上

4 u

6 u^{2} - 17 u + 5 + 4 u = - 4 u + 4 u

化简

6 u^{2} - 13 u + 5 = 0

6 u^{2} - 13 u + 5 = 0

使用求根公式求解

6 u^{2} - 13 u + 5 = 0

二次方程求根公式:

若

a = 6, b = - 13, c = 5

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 5}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 5}{2 \cdot 6}

(- 13)^{2} - 4 \cdot 6 \cdot 5 = 7

(- 13)^{2} - 4 \cdot 6 \cdot 5

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 13)^{2} = 1 3^{2}

= 1 3^{2} - 4 \cdot 6 \cdot 5

数字相乘:

4 \cdot 6 \cdot 5 = 120

= 1 3^{2} - 120

1 3^{2} = 169

= 169 - 120

数字相减:

169 - 120 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 13 ) \pm 7}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 13 ) + 7}{2 \cdot 6}, u_{2} = \frac{- ( - 13 ) - 7}{2 \cdot 6}

u = \frac{- ( - 13 ) + 7}{2 \cdot 6} : \frac{5}{3}

\frac{- ( - 13 ) + 7}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{13 + 7}{2 \cdot 6}

数字相加:

13 + 7 = 20

= \frac{20}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{20}{12}

约分:

4

= \frac{5}{3}

u = \frac{- ( - 13 ) - 7}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 13 ) - 7}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{13 - 7}{2 \cdot 6}

数字相减:

13 - 7 = 6

= \frac{6}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{6}{12}

约分:

6

= \frac{1}{2}

二次方程组的解是:

u = \frac{5}{3}, u = \frac{1}{2}

u = sin (θ)

代回

sin (θ) = \frac{5}{3}, sin (θ) = \frac{1}{2}

sin (θ) = \frac{5}{3}, sin (θ) = \frac{1}{2}

sin (θ) = \frac{5}{3} :

无解

sin (θ) = \frac{5}{3}

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 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}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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

合并所有解

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

作图

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