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受欢迎的三角函数 >

1+cos(2θ)=3sin(θ)

解答

1 + cos (2 θ) = 3 sin (θ)

解答

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

+1

度数

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

1 + cos (2 θ) = 3 sin (θ)

两边减去

3 sin (θ)

1 + cos (2 θ) - 3 sin (θ) = 0

使用三角恒等式改写

1 + cos (2 θ) - 3 sin (θ)

使用倍角公式:

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

= 1 + 1 - 2 sin^{2} (θ) - 3 sin (θ)

化简

= - 2 sin^{2} (θ) - 3 sin (θ) + 2

2 - 2 sin^{2} (θ) - 3 sin (θ) = 0

用替代法求解

2 - 2 sin^{2} (θ) - 3 sin (θ) = 0

令

: sin (θ) = u

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

2 - 2 u^{2} - 3 u = 0 : u = - 2, u = \frac{1}{2}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = - 3, c = 2

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

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

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

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

数字相乘:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数字相加:

9 + 16 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, u_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

u = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 + 5}{- 2 \cdot 2}

数字相加:

3 + 5 = 8

= \frac{8}{- 2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{8}{- 4}

使用分式法则:

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

= - \frac{8}{4}

数字相除:

\frac{8}{4} = 2

= - 2

u = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

去除括号:

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

= \frac{3 - 5}{- 2 \cdot 2}

数字相减:

3 - 5 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

使用分式法则:

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

= \frac{2}{4}

约分:

2

= \frac{1}{2}

二次方程组的解是:

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

u = sin (θ)

代回

sin (θ) = - 2, sin (θ) = \frac{1}{2}

sin (θ) = - 2, sin (θ) = \frac{1}{2}

sin (θ) = - 2 :

无解

sin (θ) = - 2

- 1 \leq sin (x) \leq 1

无解

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

合并所有解

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

作图

流行的例子

2cos(x)sin(x)-sqrt(3)sin(x)=0

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

cos(θ)-sin(θ)=cot(θ)cos(θ)

cos (θ) - sin (θ) = cot (θ) cos (θ)

-2tan(x)+1+2tan^2(x)=tan^2(x)

- 2 tan (x) + 1 + 2 tan^{2} (x) = tan^{2} (x)

(sin(a))/(50)=(sin(25))/(35)

\frac{sin ( a )}{50} = \frac{sin ( 2 5 ^{\circ} )}{35}

sin(2x+64)=(sqrt(3))/2

sin (2 x + 64) = \frac{3}{2}