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受欢迎的三角函数 >

2cos(2θ)-3sin(θ)+2=sin(θ)+5

解答

2 cos (2 θ) - 3 sin (θ) + 2 = sin (θ) + 5

解答

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

+1

度数

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos (2 θ) - 3 sin (θ) + 2 = sin (θ) + 5

两边减去

sin (θ) + 5

2 cos (2 θ) - 4 sin (θ) - 3 = 0

使用三角恒等式改写

- 3 + 2 cos (2 θ) - 4 sin (θ)

使用倍角公式:

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

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

化简

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

- 3 + 2 (1 - 2 sin^{2} (θ)) - 4 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)

= - 3 + 2 - 4 sin^{2} (θ) - 4 sin (θ)

数字相加/相减:

- 3 + 2 = - 1

= - 4 sin^{2} (θ) - 4 sin (θ) - 1

= - 4 sin^{2} (θ) - 4 sin (θ) - 1

- 1 - 4 sin (θ) - 4 sin^{2} (θ) = 0

用替代法求解

- 1 - 4 sin (θ) - 4 sin^{2} (θ) = 0

令

: sin (θ) = u

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

- 1 - 4 u - 4 u^{2} = 0 : u = - \frac{1}{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

(- 4)^{2} - 4 (- 4) (- 1) = 0

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 4^{2} - 4 \cdot 4 \cdot 1

数字相乘:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数字相减:

16 - 16 = 0

= 0

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

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

\frac{- ( - 4 )}{2 ( - 4 )} = - \frac{1}{2}

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

去除括号:

(- a) = - a, - (- a) = a

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = - \frac{1}{2}

二次方程组的解是:

u = - \frac{1}{2}

u = sin (θ)

代回

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2}

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

sin (θ) = - \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}

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

合并所有解

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

作图

流行的例子

3tan(C)+sqrt(8)=0

3 tan (C) + 8 = 0

cos (x) = - 0.8

8sin(x)cos(x)-3cos(x)=0

8 sin (x) cos (x) - 3 cos (x) = 0

sin(x)+2=3,0<x<360

sin (x) + 2 = 3, 0^{\circ} < x < 36 0^{\circ}

cos (\frac{2 x}{3}) = 0