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受欢迎的三角函数 >

2cos(2θ)-2=-3sin(θ)

解答

2 cos (2 θ) - 2 = - 3 sin (θ)

解答

θ = 2 πn, θ = π + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

+1

度数

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 48.59037 \dots^{\circ} + 36 0^{\circ} n, θ = 131.40962 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (2 θ) - 2 = - 3 sin (θ)

两边减去

- 3 sin (θ)

2 cos (2 θ) - 2 + 3 sin (θ) = 0

使用三角恒等式改写

- 2 + 2 cos (2 θ) + 3 sin (θ)

使用倍角公式:

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

= - 2 + 2 (1 - 2 sin^{2} (θ)) + 3 sin (θ)

化简

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

- 2 + 2 (1 - 2 sin^{2} (θ)) + 3 sin (θ)

乘开

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

2 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = 2 sin^{2} (θ)

= 2 \cdot 1 - 2 \cdot 2 sin^{2} (θ)

化简

2 \cdot 1 - 2 \cdot 2 sin^{2} (θ) : 2 - 4 sin^{2} (θ)

2 \cdot 1 - 2 \cdot 2 sin^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (θ)

数字相乘:

2 \cdot 2 = 4

= 2 - 4 sin^{2} (θ)

= 2 - 4 sin^{2} (θ)

= - 2 + 2 - 4 sin^{2} (θ) + 3 sin (θ)

- 2 + 2 = 0

= 3 sin (θ) - 4 sin^{2} (θ)

= 3 sin (θ) - 4 sin^{2} (θ)

3 sin (θ) - 4 sin^{2} (θ) = 0

用替代法求解

3 sin (θ) - 4 sin^{2} (θ) = 0

令

: sin (θ) = u

3 u - 4 u^{2} = 0

3 u - 4 u^{2} = 0 : u = 0, u = \frac{3}{4}

3 u - 4 u^{2} = 0

改写成标准形式

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

- 4 u^{2} + 3 u = 0

使用求根公式求解

- 4 u^{2} + 3 u = 0

二次方程求根公式:

若

a = - 4, b = 3, c = 0

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

3^{2} - 4 (- 4) \cdot 0 = 3

3^{2} - 4 (- 4) \cdot 0

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 4 \cdot 0

使用法则

0 \cdot a = 0

= 3^{2} + 0

3^{2} + 0 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 3

u_{1, 2} = \frac{- 3 \pm 3}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 3 + 3}{2 ( - 4 )}, u_{2} = \frac{- 3 - 3}{2 ( - 4 )}

u = \frac{- 3 + 3}{2 ( - 4 )} : 0

\frac{- 3 + 3}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 3 + 3}{- 2 \cdot 4}

数字相加/相减:

- 3 + 3 = 0

= \frac{0}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{0}{- 8}

使用分式法则:

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

= - \frac{0}{8}

使用法则

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 3 - 3}{2 ( - 4 )} : \frac{3}{4}

\frac{- 3 - 3}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 3 - 3}{- 2 \cdot 4}

数字相减:

- 3 - 3 = - 6

= \frac{- 6}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 6}{- 8}

使用分式法则:

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

= \frac{6}{8}

约分:

2

= \frac{3}{4}

二次方程组的解是:

u = 0, u = \frac{3}{4}

u = sin (θ)

代回

sin (θ) = 0, sin (θ) = \frac{3}{4}

sin (θ) = 0, sin (θ) = \frac{3}{4}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

sin (θ) = 0

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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (θ) = \frac{3}{4} : θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

sin (θ) = \frac{3}{4}

使用反三角函数性质

sin (θ) = \frac{3}{4}

sin (θ) = \frac{3}{4}

的通解

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

θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

合并所有解

θ = 2 πn, θ = π + 2 πn, θ = arcsin (\frac{3}{4}) + 2 πn, θ = π - arcsin (\frac{3}{4}) + 2 πn

以小数形式表示解

θ = 2 πn, θ = π + 2 πn, θ = 0.84806 \dots + 2 πn, θ = π - 0.84806 \dots + 2 πn

作图

流行的例子

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sin(pi/6+x)+sin(pi/6-x)= 1/2

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