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受欢迎的三角函数 >

2cos(2θ)+7sin(θ)-2=-2

解答

2 cos (2 θ) + 7 sin (θ) - 2 = - 2

解答

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

+1

度数

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

求解步骤

2 cos (2 θ) + 7 sin (θ) - 2 = - 2

两边减去

- 2

2 cos (2 θ) + 7 sin (θ) = 0

使用三角恒等式改写

2 cos (2 θ) + 7 sin (θ)

使用倍角公式:

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

= 2 (1 - 2 sin^{2} (θ)) + 7 sin (θ)

(1 - 2 sin^{2} (θ)) \cdot 2 + 7 sin (θ) = 0

用替代法求解

(1 - 2 sin^{2} (θ)) \cdot 2 + 7 sin (θ) = 0

令

: sin (θ) = u

(1 - 2 u^{2}) \cdot 2 + 7 u = 0

(1 - 2 u^{2}) \cdot 2 + 7 u = 0 : u = - \frac{1}{4}, u = 2

(1 - 2 u^{2}) \cdot 2 + 7 u = 0

展开

(1 - 2 u^{2}) \cdot 2 + 7 u : 2 - 4 u^{2} + 7 u

(1 - 2 u^{2}) \cdot 2 + 7 u

= 2 (1 - 2 u^{2}) + 7 u

乘开

2 (1 - 2 u^{2}) : 2 - 4 u^{2}

2 (1 - 2 u^{2})

使用分配律:

a (b - c) = ab - a c

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

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

化简

2 \cdot 1 - 2 \cdot 2 u^{2} : 2 - 4 u^{2}

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

数字相乘:

2 \cdot 1 = 2

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

数字相乘:

2 \cdot 2 = 4

= 2 - 4 u^{2}

= 2 - 4 u^{2}

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 4, b = 7, c = 2

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

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

7^{2} - 4 (- 4) \cdot 2 = 9

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 7^{2} + 32

7^{2} = 49

= 49 + 32

数字相加:

49 + 32 = 81

= 81

因式分解数字:

81 = 9^{2}

= 9^{2}

使用根式运算法则:

n a^{n} = a

9^{2} = 9

= 9

u_{1, 2} = \frac{- 7 \pm 9}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 7 + 9}{2 ( - 4 )}, u_{2} = \frac{- 7 - 9}{2 ( - 4 )}

u = \frac{- 7 + 9}{2 ( - 4 )} : - \frac{1}{4}

\frac{- 7 + 9}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 7 + 9}{- 2 \cdot 4}

数字相加/相减:

- 7 + 9 = 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}

u = \frac{- 7 - 9}{2 ( - 4 )} : 2

\frac{- 7 - 9}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 7 - 9}{- 2 \cdot 4}

数字相减:

- 7 - 9 = - 16

= \frac{- 16}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 16}{- 8}

使用分式法则:

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

= \frac{16}{8}

数字相除:

\frac{16}{8} = 2

= 2

二次方程组的解是:

u = - \frac{1}{4}, u = 2

u = sin (θ)

代回

sin (θ) = - \frac{1}{4}, sin (θ) = 2

sin (θ) = - \frac{1}{4}, sin (θ) = 2

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

sin (θ) = - \frac{1}{4}

使用反三角函数性质

sin (θ) = - \frac{1}{4}

sin (θ) = - \frac{1}{4}

的通解

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

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

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

sin (θ) = 2 :

无解

sin (θ) = 2

- 1 \leq sin (x) \leq 1

无解

合并所有解

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

以小数形式表示解

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

作图

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