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受欢迎的三角函数 >

-3cos(2θ)+3=5sin(θ)

解答

- 3 cos (2 θ) + 3 = 5 sin (θ)

解答

θ = 0.98511 \dots + 2 πn, θ = π - 0.98511 \dots + 2 πn, θ = 2 πn, θ = π + 2 πn

+1

度数

θ = 56.44269 \dots^{\circ} + 36 0^{\circ} n, θ = 123.55730 \dots^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

- 3 cos (2 θ) + 3 = 5 sin (θ)

两边减去

5 sin (θ)

- 3 cos (2 θ) + 3 - 5 sin (θ) = 0

使用三角恒等式改写

3 - 3 cos (2 θ) - 5 sin (θ)

使用倍角公式:

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

= 3 - 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ)

化简

3 - 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ) : 6 sin^{2} (θ) - 5 sin (θ)

3 - 3 (1 - 2 sin^{2} (θ)) - 5 sin (θ)

乘开

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

- 3 (1 - 2 sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = 2 sin^{2} (θ)

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

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

化简

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

- 3 \cdot 1 + 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 \cdot 2 sin^{2} (θ)

数字相乘:

3 \cdot 2 = 6

= - 3 + 6 sin^{2} (θ)

= - 3 + 6 sin^{2} (θ)

= 3 - 3 + 6 sin^{2} (θ) - 5 sin (θ)

3 - 3 = 0

= 6 sin^{2} (θ) - 5 sin (θ)

= 6 sin^{2} (θ) - 5 sin (θ)

- 5 sin (θ) + 6 sin^{2} (θ) = 0

用替代法求解

- 5 sin (θ) + 6 sin^{2} (θ) = 0

令

: sin (θ) = u

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

- 5 u + 6 u^{2} = 0 : u = \frac{5}{6}, u = 0

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

改写成标准形式

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

6 u^{2} - 5 u = 0

使用求根公式求解

6 u^{2} - 5 u = 0

二次方程求根公式:

若

a = 6, b = - 5, c = 0

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 \cdot 0 = 5

(- 5)^{2} - 4 \cdot 6 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 6 \cdot 0

使用法则

0 \cdot a = 0

= 5^{2} - 0

5^{2} - 0 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 5

u_{1, 2} = \frac{- ( - 5 ) \pm 5}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- ( - 5 ) + 5}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 5}{2 \cdot 6}

u = \frac{- ( - 5 ) + 5}{2 \cdot 6} : \frac{5}{6}

\frac{- ( - 5 ) + 5}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{5 + 5}{2 \cdot 6}

数字相加:

5 + 5 = 10

= \frac{10}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{10}{12}

约分:

2

= \frac{5}{6}

u = \frac{- ( - 5 ) - 5}{2 \cdot 6} : 0

\frac{- ( - 5 ) - 5}{2 \cdot 6}

使用法则

- (- a) = a

= \frac{5 - 5}{2 \cdot 6}

数字相减:

5 - 5 = 0

= \frac{0}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{0}{12}

使用法则

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

= 0

二次方程组的解是:

u = \frac{5}{6}, u = 0

u = sin (θ)

代回

sin (θ) = \frac{5}{6}, sin (θ) = 0

sin (θ) = \frac{5}{6}, sin (θ) = 0

sin (θ) = \frac{5}{6} : θ = arcsin (\frac{5}{6}) + 2 πn, θ = π - arcsin (\frac{5}{6}) + 2 πn

sin (θ) = \frac{5}{6}

使用反三角函数性质

sin (θ) = \frac{5}{6}

sin (θ) = \frac{5}{6}

的通解

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

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

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

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

合并所有解

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

以小数形式表示解

θ = 0.98511 \dots + 2 πn, θ = π - 0.98511 \dots + 2 πn, θ = 2 πn, θ = π + 2 πn

作图

流行的例子

12sin^2(x)+18sin(x)+6=0

12 sin^{2} (x) + 18 sin (x) + 6 = 0

csc^2(θ)+cot(θ)=0

csc^{2} (θ) + cot (θ) = 0

sin^2(θ)-2sin(θ)-3=0

sin^{2} (θ) - 2 sin (θ) - 3 = 0

5 tan (θ) - 4 = 0

0 = cos (\frac{x}{2})