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受欢迎的三角函数 >

-8sin(x)+8cos(2x)=0

解答

- 8 sin (x) + 8 cos (2 x) = 0

解答

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 27 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

- 8 sin (x) + 8 cos (2 x) = 0

使用三角恒等式改写

8 cos (2 x) - 8 sin (x)

使用倍角公式:

cos (2 x) = 1 - 2 sin^{2} (x)

= 8 (1 - 2 sin^{2} (x)) - 8 sin (x)

(1 - 2 sin^{2} (x)) \cdot 8 - 8 sin (x) = 0

用替代法求解

(1 - 2 sin^{2} (x)) \cdot 8 - 8 sin (x) = 0

令

: sin (x) = u

(1 - 2 u^{2}) \cdot 8 - 8 u = 0

(1 - 2 u^{2}) \cdot 8 - 8 u = 0 : u = - 1, u = \frac{1}{2}

(1 - 2 u^{2}) \cdot 8 - 8 u = 0

展开

(1 - 2 u^{2}) \cdot 8 - 8 u : 8 - 16 u^{2} - 8 u

(1 - 2 u^{2}) \cdot 8 - 8 u

= 8 (1 - 2 u^{2}) - 8 u

乘开

8 (1 - 2 u^{2}) : 8 - 16 u^{2}

8 (1 - 2 u^{2})

使用分配律:

a (b - c) = ab - a c

a = 8, b = 1, c = 2 u^{2}

= 8 \cdot 1 - 8 \cdot 2 u^{2}

化简

8 \cdot 1 - 8 \cdot 2 u^{2} : 8 - 16 u^{2}

8 \cdot 1 - 8 \cdot 2 u^{2}

数字相乘:

8 \cdot 1 = 8

= 8 - 8 \cdot 2 u^{2}

数字相乘:

8 \cdot 2 = 16

= 8 - 16 u^{2}

= 8 - 16 u^{2}

= 8 - 16 u^{2} - 8 u

8 - 16 u^{2} - 8 u = 0

改写成标准形式

a x^{2} + b x + c = 0

- 16 u^{2} - 8 u + 8 = 0

使用求根公式求解

- 16 u^{2} - 8 u + 8 = 0

二次方程求根公式:

若

a = - 16, b = - 8, c = 8

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 8}{2 ( - 16 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 16 ) \cdot 8}{2 ( - 16 )}

(- 8)^{2} - 4 (- 16) \cdot 8 = 24

(- 8)^{2} - 4 (- 16) \cdot 8

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 16 \cdot 8

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 16 \cdot 8

数字相乘:

4 \cdot 16 \cdot 8 = 512

= 8^{2} + 512

8^{2} = 64

= 64 + 512

数字相加:

64 + 512 = 576

= 576

因式分解数字:

576 = 2 4^{2}

= 2 4^{2}

使用根式运算法则:

n a^{n} = a

2 4^{2} = 24

= 24

u_{1, 2} = \frac{- ( - 8 ) \pm 24}{2 ( - 16 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 24}{2 ( - 16 )}, u_{2} = \frac{- ( - 8 ) - 24}{2 ( - 16 )}

u = \frac{- ( - 8 ) + 24}{2 ( - 16 )} : - 1

\frac{- ( - 8 ) + 24}{2 ( - 16 )}

去除括号:

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

= \frac{8 + 24}{- 2 \cdot 16}

数字相加:

8 + 24 = 32

= \frac{32}{- 2 \cdot 16}

数字相乘:

2 \cdot 16 = 32

= \frac{32}{- 32}

使用分式法则:

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

= - \frac{32}{32}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 8 ) - 24}{2 ( - 16 )} : \frac{1}{2}

\frac{- ( - 8 ) - 24}{2 ( - 16 )}

去除括号:

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

= \frac{8 - 24}{- 2 \cdot 16}

数字相减:

8 - 24 = - 16

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

数字相乘:

2 \cdot 16 = 32

= \frac{- 16}{- 32}

使用分式法则:

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

= \frac{16}{32}

约分:

16

= \frac{1}{2}

二次方程组的解是:

u = - 1, u = \frac{1}{2}

u = sin (x)

代回

sin (x) = - 1, sin (x) = \frac{1}{2}

sin (x) = - 1, sin (x) = \frac{1}{2}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

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

sin (x) = \frac{1}{2}

sin (x) = \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}

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

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

合并所有解

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

作图

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