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受欢迎的三角函数 >

8sin^2(θ)-17sin(θ)+9=0

解答

8 sin^{2} (θ) - 17 sin (θ) + 9 = 0

解答

θ = \frac{π}{2} + 2 πn

+1

度数

θ = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

8 sin^{2} (θ) - 17 sin (θ) + 9 = 0

用替代法求解

8 sin^{2} (θ) - 17 sin (θ) + 9 = 0

令

: sin (θ) = u

8 u^{2} - 17 u + 9 = 0

8 u^{2} - 17 u + 9 = 0 : u = \frac{9}{8}, u = 1

8 u^{2} - 17 u + 9 = 0

使用求根公式求解

8 u^{2} - 17 u + 9 = 0

二次方程求根公式:

若

a = 8, b = - 17, c = 9

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 8 \cdot 9}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 17 ) \pm ( - 17 ) ^{2} - 4 \cdot 8 \cdot 9}{2 \cdot 8}

(- 17)^{2} - 4 \cdot 8 \cdot 9 = 1

(- 17)^{2} - 4 \cdot 8 \cdot 9

使用指数法则:

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

若

n

是偶数

(- 17)^{2} = 1 7^{2}

= 1 7^{2} - 4 \cdot 8 \cdot 9

数字相乘:

4 \cdot 8 \cdot 9 = 288

= 1 7^{2} - 288

1 7^{2} = 289

= 289 - 288

数字相减:

289 - 288 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- ( - 17 ) \pm 1}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- ( - 17 ) + 1}{2 \cdot 8}, u_{2} = \frac{- ( - 17 ) - 1}{2 \cdot 8}

u = \frac{- ( - 17 ) + 1}{2 \cdot 8} : \frac{9}{8}

\frac{- ( - 17 ) + 1}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{17 + 1}{2 \cdot 8}

数字相加:

17 + 1 = 18

= \frac{18}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{18}{16}

约分:

2

= \frac{9}{8}

u = \frac{- ( - 17 ) - 1}{2 \cdot 8} : 1

\frac{- ( - 17 ) - 1}{2 \cdot 8}

使用法则

- (- a) = a

= \frac{17 - 1}{2 \cdot 8}

数字相减:

17 - 1 = 16

= \frac{16}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{16}{16}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = \frac{9}{8}, u = 1

u = sin (θ)

代回

sin (θ) = \frac{9}{8}, sin (θ) = 1

sin (θ) = \frac{9}{8}, sin (θ) = 1

sin (θ) = \frac{9}{8} :

无解

sin (θ) = \frac{9}{8}

- 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

流行的例子

sin^2(x)=2cos^2(x)

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

9sin(2θ)-2sin(θ)=0

9 sin (2 θ) - 2 sin (θ) = 0

3 csc (θ) - 4 = 0

cos(2x)-sin(3x)=0

cos (2 x) - sin (3 x) = 0

sec(θ)=2cos(θ)

sec (θ) = 2 cos (θ)