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受欢迎的三角函数 >

4(sin(x))^2-15sin(x)-4=0

解答

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

解答

x = - 0.25268 \dots + 2 πn, x = π + 0.25268 \dots + 2 πn

+1

度数

x = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, x = 194.47751 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

用替代法求解

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

令

: sin (x) = u

4 u^{2} - 15 u - 4 = 0

4 u^{2} - 15 u - 4 = 0 : u = 4, u = - \frac{1}{4}

4 u^{2} - 15 u - 4 = 0

使用求根公式求解

4 u^{2} - 15 u - 4 = 0

二次方程求根公式:

若

a = 4, b = - 15, c = - 4

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

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

(- 15)^{2} - 4 \cdot 4 (- 4) = 17

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

使用法则

- (- a) = a

= (- 15)^{2} + 4 \cdot 4 \cdot 4

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 4 \cdot 4

数字相乘:

4 \cdot 4 \cdot 4 = 64

= 1 5^{2} + 64

1 5^{2} = 225

= 225 + 64

数字相加:

225 + 64 = 289

= 289

因式分解数字:

289 = 1 7^{2}

= 1 7^{2}

使用根式运算法则:

n a^{n} = a

1 7^{2} = 17

= 17

u_{1, 2} = \frac{- ( - 15 ) \pm 17}{2 \cdot 4}

将解分隔开

u_{1} = \frac{- ( - 15 ) + 17}{2 \cdot 4}, u_{2} = \frac{- ( - 15 ) - 17}{2 \cdot 4}

u = \frac{- ( - 15 ) + 17}{2 \cdot 4} : 4

\frac{- ( - 15 ) + 17}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{15 + 17}{2 \cdot 4}

数字相加:

15 + 17 = 32

= \frac{32}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{32}{8}

数字相除:

\frac{32}{8} = 4

= 4

u = \frac{- ( - 15 ) - 17}{2 \cdot 4} : - \frac{1}{4}

\frac{- ( - 15 ) - 17}{2 \cdot 4}

使用法则

- (- a) = a

= \frac{15 - 17}{2 \cdot 4}

数字相减:

15 - 17 = - 2

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 2}{8}

使用分式法则:

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

= - \frac{2}{8}

约分:

2

= - \frac{1}{4}

二次方程组的解是:

u = 4, u = - \frac{1}{4}

u = sin (x)

代回

sin (x) = 4, sin (x) = - \frac{1}{4}

sin (x) = 4, sin (x) = - \frac{1}{4}

sin (x) = 4 :

无解

sin (x) = 4

- 1 \leq sin (x) \leq 1

无解

sin (x) = - \frac{1}{4} : x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

sin (x) = - \frac{1}{4}

使用反三角函数性质

sin (x) = - \frac{1}{4}

sin (x) = - \frac{1}{4}

的通解

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

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

合并所有解

x = arcsin (- \frac{1}{4}) + 2 πn, x = π + arcsin (\frac{1}{4}) + 2 πn

以小数形式表示解

x = - 0.25268 \dots + 2 πn, x = π + 0.25268 \dots + 2 πn

作图

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