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受欢迎的三角函数 >

4cos^2(x)+5sin(x)=3

解答

4 cos^{2} (x) + 5 sin (x) = 3

解答

x = - 0.17630 \dots + 2 πn, x = π + 0.17630 \dots + 2 πn

+1

度数

x = - 10.10138 \dots^{\circ} + 36 0^{\circ} n, x = 190.10138 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos^{2} (x) + 5 sin (x) = 3

两边减去

3

4 cos^{2} (x) + 5 sin (x) - 3 = 0

使用三角恒等式改写

- 3 + 4 cos^{2} (x) + 5 sin (x)

使用毕达哥拉斯恒等式:

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

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

= - 3 + 4 (1 - sin^{2} (x)) + 5 sin (x)

化简

- 3 + 4 (1 - sin^{2} (x)) + 5 sin (x) : 5 sin (x) - 4 sin^{2} (x) + 1

- 3 + 4 (1 - sin^{2} (x)) + 5 sin (x)

乘开

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= - 3 + 4 - 4 sin^{2} (x) + 5 sin (x)

数字相加/相减:

- 3 + 4 = 1

= 5 sin (x) - 4 sin^{2} (x) + 1

= 5 sin (x) - 4 sin^{2} (x) + 1

1 - 4 sin^{2} (x) + 5 sin (x) = 0

用替代法求解

1 - 4 sin^{2} (x) + 5 sin (x) = 0

令

: sin (x) = u

1 - 4 u^{2} + 5 u = 0

1 - 4 u^{2} + 5 u = 0 : u = - \frac{- 5 + 41}{8}, u = \frac{5 + 41}{8}

1 - 4 u^{2} + 5 u = 0

改写成标准形式

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

- 4 u^{2} + 5 u + 1 = 0

使用求根公式求解

- 4 u^{2} + 5 u + 1 = 0

二次方程求根公式:

若

a = - 4, b = 5, c = 1

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 4 ) \cdot 1}{2 ( - 4 )}

5^{2} - 4 (- 4) \cdot 1 = 41

5^{2} - 4 (- 4) \cdot 1

使用法则

- (- a) = a

= 5^{2} + 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 5^{2} + 16

5^{2} = 25

= 25 + 16

数字相加:

25 + 16 = 41

= 41

u_{1, 2} = \frac{- 5 \pm 41}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 5 + 41}{2 ( - 4 )}, u_{2} = \frac{- 5 - 41}{2 ( - 4 )}

u = \frac{- 5 + 41}{2 ( - 4 )} : - \frac{- 5 + 41}{8}

\frac{- 5 + 41}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 5 + 41}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 5 + 41}{- 8}

使用分式法则:

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

= - \frac{- 5 + 41}{8}

u = \frac{- 5 - 41}{2 ( - 4 )} : \frac{5 + 41}{8}

\frac{- 5 - 41}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 5 - 41}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 5 - 41}{- 8}

使用分式法则:

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

- 5 - 41 = - (5 + 41)

= \frac{5 + 41}{8}

二次方程组的解是:

u = - \frac{- 5 + 41}{8}, u = \frac{5 + 41}{8}

u = sin (x)

代回

sin (x) = - \frac{- 5 + 41}{8}, sin (x) = \frac{5 + 41}{8}

sin (x) = - \frac{- 5 + 41}{8}, sin (x) = \frac{5 + 41}{8}

sin (x) = - \frac{- 5 + 41}{8} : x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

sin (x) = - \frac{- 5 + 41}{8}

使用反三角函数性质

sin (x) = - \frac{- 5 + 41}{8}

sin (x) = - \frac{- 5 + 41}{8}

的通解

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

x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

sin (x) = \frac{5 + 41}{8} :

无解

sin (x) = \frac{5 + 41}{8}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = arcsin (- \frac{- 5 + 41}{8}) + 2 πn, x = π + arcsin (\frac{- 5 + 41}{8}) + 2 πn

以小数形式表示解

x = - 0.17630 \dots + 2 πn, x = π + 0.17630 \dots + 2 πn

作图

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