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受欢迎的三角函数 >

7sin^2(θ)-15sin(θ)+8=0

解答

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

解答

θ = \frac{π}{2} + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

用替代法求解

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

令

: sin (θ) = u

7 u^{2} - 15 u + 8 = 0

7 u^{2} - 15 u + 8 = 0 : u = \frac{8}{7}, u = 1

7 u^{2} - 15 u + 8 = 0

使用求根公式求解

7 u^{2} - 15 u + 8 = 0

二次方程求根公式:

若

a = 7, b = - 15, c = 8

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 7 \cdot 8}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 7 \cdot 8}{2 \cdot 7}

(- 15)^{2} - 4 \cdot 7 \cdot 8 = 1

(- 15)^{2} - 4 \cdot 7 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 15)^{2} = 1 5^{2}

= 1 5^{2} - 4 \cdot 7 \cdot 8

数字相乘:

4 \cdot 7 \cdot 8 = 224

= 1 5^{2} - 224

1 5^{2} = 225

= 225 - 224

数字相减:

225 - 224 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 15 ) \pm 1}{2 \cdot 7}

将解分隔开

u_{1} = \frac{- ( - 15 ) + 1}{2 \cdot 7}, u_{2} = \frac{- ( - 15 ) - 1}{2 \cdot 7}

u = \frac{- ( - 15 ) + 1}{2 \cdot 7} : \frac{8}{7}

\frac{- ( - 15 ) + 1}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{15 + 1}{2 \cdot 7}

数字相加:

15 + 1 = 16

= \frac{16}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{16}{14}

约分:

2

= \frac{8}{7}

u = \frac{- ( - 15 ) - 1}{2 \cdot 7} : 1

\frac{- ( - 15 ) - 1}{2 \cdot 7}

使用法则

- (- a) = a

= \frac{15 - 1}{2 \cdot 7}

数字相减:

15 - 1 = 14

= \frac{14}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{14}{14}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = \frac{8}{7}, u = 1

u = sin (θ)

代回

sin (θ) = \frac{8}{7}, sin (θ) = 1

sin (θ) = \frac{8}{7}, sin (θ) = 1

sin (θ) = \frac{8}{7} :

无解

sin (θ) = \frac{8}{7}

- 1 \leq sin (x) \leq 1

无解

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

合并所有解

θ = \frac{π}{2} + 2 πn

作图

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