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受欢迎的三角函数 >

3cos(2x)+7sin(x)-5=0

解答

3 cos (2 x) + 7 sin (x) - 5 = 0

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 41.81031 \dots^{\circ} + 36 0^{\circ} n, x = 138.18968 \dots^{\circ} + 36 0^{\circ} n

求解步骤

3 cos (2 x) + 7 sin (x) - 5 = 0

使用三角恒等式改写

- 5 + 3 cos (2 x) + 7 sin (x)

使用倍角公式:

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

= - 5 + 3 (1 - 2 sin^{2} (x)) + 7 sin (x)

化简

- 5 + 3 (1 - 2 sin^{2} (x)) + 7 sin (x) : 7 sin (x) - 6 sin^{2} (x) - 2

- 5 + 3 (1 - 2 sin^{2} (x)) + 7 sin (x)

乘开

3 (1 - 2 sin^{2} (x)) : 3 - 6 sin^{2} (x)

3 (1 - 2 sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = 2 sin^{2} (x)

= 3 \cdot 1 - 3 \cdot 2 sin^{2} (x)

化简

3 \cdot 1 - 3 \cdot 2 sin^{2} (x) : 3 - 6 sin^{2} (x)

3 \cdot 1 - 3 \cdot 2 sin^{2} (x)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 \cdot 2 sin^{2} (x)

数字相乘:

3 \cdot 2 = 6

= 3 - 6 sin^{2} (x)

= 3 - 6 sin^{2} (x)

= - 5 + 3 - 6 sin^{2} (x) + 7 sin (x)

数字相加/相减:

- 5 + 3 = - 2

= 7 sin (x) - 6 sin^{2} (x) - 2

= 7 sin (x) - 6 sin^{2} (x) - 2

- 2 - 6 sin^{2} (x) + 7 sin (x) = 0

用替代法求解

- 2 - 6 sin^{2} (x) + 7 sin (x) = 0

令

: sin (x) = u

- 2 - 6 u^{2} + 7 u = 0

- 2 - 6 u^{2} + 7 u = 0 : u = \frac{1}{2}, u = \frac{2}{3}

- 2 - 6 u^{2} + 7 u = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 6, b = 7, c = - 2

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 6 ) ( - 2 )}{2 ( - 6 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 6 ) ( - 2 )}{2 ( - 6 )}

7^{2} - 4 (- 6) (- 2) = 1

7^{2} - 4 (- 6) (- 2)

使用法则

- (- a) = a

= 7^{2} - 4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

数字相减:

49 - 48 = 1

= 1

使用法则

1 = 1

= 1

u_{1, 2} = \frac{- 7 \pm 1}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 7 + 1}{2 ( - 6 )}, u_{2} = \frac{- 7 - 1}{2 ( - 6 )}

u = \frac{- 7 + 1}{2 ( - 6 )} : \frac{1}{2}

\frac{- 7 + 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 7 + 1}{- 2 \cdot 6}

数字相加/相减:

- 7 + 1 = - 6

= \frac{- 6}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{- 12}

使用分式法则:

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

= \frac{6}{12}

约分:

6

= \frac{1}{2}

u = \frac{- 7 - 1}{2 ( - 6 )} : \frac{2}{3}

\frac{- 7 - 1}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 7 - 1}{- 2 \cdot 6}

数字相减:

- 7 - 1 = - 8

= \frac{- 8}{- 2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{- 8}{- 12}

使用分式法则:

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

= \frac{8}{12}

约分:

4

= \frac{2}{3}

二次方程组的解是:

u = \frac{1}{2}, u = \frac{2}{3}

u = sin (x)

代回

sin (x) = \frac{1}{2}, sin (x) = \frac{2}{3}

sin (x) = \frac{1}{2}, sin (x) = \frac{2}{3}

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

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

sin (x) = \frac{2}{3}

使用反三角函数性质

sin (x) = \frac{2}{3}

sin (x) = \frac{2}{3}

的通解

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

x = arcsin (\frac{2}{3}) + 2 πn, x = π - arcsin (\frac{2}{3}) + 2 πn

x = arcsin (\frac{2}{3}) + 2 πn, x = π - arcsin (\frac{2}{3}) + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arcsin (\frac{2}{3}) + 2 πn, x = π - arcsin (\frac{2}{3}) + 2 πn

以小数形式表示解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

作图

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