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受欢迎的三角函数 >

-4sin^2(θ)-7sin(θ)+7=0

解答

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

解答

θ = 0.79102 \dots + 2 πn, θ = π - 0.79102 \dots + 2 πn

+1

度数

θ = 45.32221 \dots^{\circ} + 36 0^{\circ} n, θ = 134.67778 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

用替代法求解

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

令

: sin (θ) = u

- 4 u^{2} - 7 u + 7 = 0

- 4 u^{2} - 7 u + 7 = 0 : u = - \frac{7 + 161}{8}, u = \frac{161 - 7}{8}

- 4 u^{2} - 7 u + 7 = 0

使用求根公式求解

- 4 u^{2} - 7 u + 7 = 0

二次方程求根公式:

若

a = - 4, b = - 7, c = 7

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

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

(- 7)^{2} - 4 (- 4) \cdot 7 = 161

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

使用法则

- (- a) = a

= (- 7)^{2} + 4 \cdot 4 \cdot 7

使用指数法则:

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

若

n

是偶数

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

= 7^{2} + 4 \cdot 4 \cdot 7

数字相乘:

4 \cdot 4 \cdot 7 = 112

= 7^{2} + 112

7^{2} = 49

= 49 + 112

数字相加:

49 + 112 = 161

= 161

u_{1, 2} = \frac{- ( - 7 ) \pm 161}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- ( - 7 ) + 161}{2 ( - 4 )}, u_{2} = \frac{- ( - 7 ) - 161}{2 ( - 4 )}

u = \frac{- ( - 7 ) + 161}{2 ( - 4 )} : - \frac{7 + 161}{8}

\frac{- ( - 7 ) + 161}{2 ( - 4 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{7 + 161}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{7 + 161}{- 8}

使用分式法则:

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

= - \frac{7 + 161}{8}

u = \frac{- ( - 7 ) - 161}{2 ( - 4 )} : \frac{161 - 7}{8}

\frac{- ( - 7 ) - 161}{2 ( - 4 )}

去除括号:

(- a) = - a, - (- a) = a

= \frac{7 - 161}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{7 - 161}{- 8}

使用分式法则:

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

7 - 161 = - (161 - 7)

= \frac{161 - 7}{8}

二次方程组的解是:

u = - \frac{7 + 161}{8}, u = \frac{161 - 7}{8}

u = sin (θ)

代回

sin (θ) = - \frac{7 + 161}{8}, sin (θ) = \frac{161 - 7}{8}

sin (θ) = - \frac{7 + 161}{8}, sin (θ) = \frac{161 - 7}{8}

sin (θ) = - \frac{7 + 161}{8} :

无解

sin (θ) = - \frac{7 + 161}{8}

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{161 - 7}{8} : θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

sin (θ) = \frac{161 - 7}{8}

使用反三角函数性质

sin (θ) = \frac{161 - 7}{8}

sin (θ) = \frac{161 - 7}{8}

的通解

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

θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

合并所有解

θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

以小数形式表示解

θ = 0.79102 \dots + 2 πn, θ = π - 0.79102 \dots + 2 πn

作图

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