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受欢迎的三角函数 >

2sin(θ)+cos(2θ)=0

解答

2 sin (θ) + cos (2 θ) = 0

解答

θ = - 0.37473 \dots + 2 πn, θ = π + 0.37473 \dots + 2 πn

+1

度数

θ = - 21.47070 \dots^{\circ} + 36 0^{\circ} n, θ = 201.47070 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (θ) + cos (2 θ) = 0

使用三角恒等式改写

cos (2 θ) + 2 sin (θ)

使用倍角公式:

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

= 1 - 2 sin^{2} (θ) + 2 sin (θ)

1 + 2 sin (θ) - 2 sin^{2} (θ) = 0

用替代法求解

1 + 2 sin (θ) - 2 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 2, b = 2, c = 1

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

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

2^{2} - 4 (- 2) \cdot 1 = 23

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

使用法则

- (- a) = a

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

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 2^{2} + 8

2^{2} = 4

= 4 + 8

数字相加:

4 + 8 = 12

= 12

12

质因数分解:

2^{2} \cdot 3

12

12

除以

212 = 6 \cdot 2

= 2 \cdot 6

6

除以

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

使用根式运算法则:

n ab = n a n b

= 3 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 23

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

将解分隔开

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

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 2 = 4

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

使用分式法则:

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

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

消掉

\frac{- 2 + 2 3}{4} : \frac{3 - 1}{2}

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

分解

- 2 + 23 : 2 (- 1 + 3)

- 2 + 23

改写为

= - 2 \cdot 1 + 23

因式分解出通项

2

= 2 (- 1 + 3)

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

约分:

2

= \frac{- 1 + 3}{2}

= - \frac{3 - 1}{2}

= - \frac{- 1 + 3}{2}

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

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

去除括号:

(- a) = - a

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

数字相乘:

2 \cdot 2 = 4

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

使用分式法则:

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

- 2 - 23 = - (2 + 23)

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

分解

2 + 23 : 2 (1 + 3)

2 + 23

改写为

= 2 \cdot 1 + 23

因式分解出通项

2

= 2 (1 + 3)

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

约分:

2

= \frac{1 + 3}{2}

二次方程组的解是:

u = - \frac{- 1 + 3}{2}, u = \frac{1 + 3}{2}

u = sin (θ)

代回

sin (θ) = - \frac{- 1 + 3}{2}, sin (θ) = \frac{1 + 3}{2}

sin (θ) = - \frac{- 1 + 3}{2}, sin (θ) = \frac{1 + 3}{2}

sin (θ) = - \frac{- 1 + 3}{2} : θ = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, θ = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

sin (θ) = - \frac{- 1 + 3}{2}

使用反三角函数性质

sin (θ) = - \frac{- 1 + 3}{2}

sin (θ) = - \frac{- 1 + 3}{2}

的通解

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

θ = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, θ = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

θ = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, θ = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

sin (θ) = \frac{1 + 3}{2} :

无解

sin (θ) = \frac{1 + 3}{2}

- 1 \leq sin (x) \leq 1

无解

合并所有解

θ = arcsin (- \frac{- 1 + 3}{2}) + 2 πn, θ = π + arcsin (\frac{- 1 + 3}{2}) + 2 πn

以小数形式表示解

θ = - 0.37473 \dots + 2 πn, θ = π + 0.37473 \dots + 2 πn

作图

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