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受欢迎的三角函数 >

4cos(2x)=-2sin(x)

解答

4 cos (2 x) = - 2 sin (x)

解答

x = - 0.63486 \dots + 2 πn, x = π + 0.63486 \dots + 2 πn, x = 1.00296 \dots + 2 πn, x = π - 1.00296 \dots + 2 πn

+1

度数

x = - 36.37519 \dots^{\circ} + 36 0^{\circ} n, x = 216.37519 \dots^{\circ} + 36 0^{\circ} n, x = 57.46577 \dots^{\circ} + 36 0^{\circ} n, x = 122.53422 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 cos (2 x) = - 2 sin (x)

两边减去

- 2 sin (x)

4 cos (2 x) + 2 sin (x) = 0

使用三角恒等式改写

2 sin (x) + 4 cos (2 x)

使用倍角公式:

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

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

(1 - 2 sin^{2} (x)) \cdot 4 + 2 sin (x) = 0

用替代法求解

(1 - 2 sin^{2} (x)) \cdot 4 + 2 sin (x) = 0

令

: sin (x) = u

(1 - 2 u^{2}) \cdot 4 + 2 u = 0

(1 - 2 u^{2}) \cdot 4 + 2 u = 0 : u = - \frac{- 1 + 33}{8}, u = \frac{1 + 33}{8}

(1 - 2 u^{2}) \cdot 4 + 2 u = 0

展开

(1 - 2 u^{2}) \cdot 4 + 2 u : 4 - 8 u^{2} + 2 u

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

= 4 (1 - 2 u^{2}) + 2 u

乘开

4 (1 - 2 u^{2}) : 4 - 8 u^{2}

4 (1 - 2 u^{2})

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = 2 u^{2}

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

化简

4 \cdot 1 - 4 \cdot 2 u^{2} : 4 - 8 u^{2}

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

数字相乘:

4 \cdot 1 = 4

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

数字相乘:

4 \cdot 2 = 8

= 4 - 8 u^{2}

= 4 - 8 u^{2}

= 4 - 8 u^{2} + 2 u

4 - 8 u^{2} + 2 u = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 8, b = 2, c = 4

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 8 ) \cdot 4}{2 ( - 8 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 8 ) \cdot 4}{2 ( - 8 )}

2^{2} - 4 (- 8) \cdot 4 = 233

2^{2} - 4 (- 8) \cdot 4

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 8 \cdot 4

数字相乘:

4 \cdot 8 \cdot 4 = 128

= 2^{2} + 128

2^{2} = 4

= 4 + 128

数字相加:

4 + 128 = 132

= 132

132

质因数分解:

2^{2} \cdot 3 \cdot 11

132

132

除以

2132 = 66 \cdot 2

= 2 \cdot 66

66

除以

266 = 33 \cdot 2

= 2 \cdot 2 \cdot 33

33

除以

333 = 11 \cdot 3

= 2 \cdot 2 \cdot 3 \cdot 11

2, 3, 11

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 3 \cdot 11

= 2^{2} \cdot 3 \cdot 11

= 2^{2} \cdot 3 \cdot 11

使用根式运算法则:

n ab = n a n b

= 2^{2} 3 \cdot 11

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3 \cdot 11

整理后得

= 233

u_{1, 2} = \frac{- 2 \pm 2 33}{2 ( - 8 )}

将解分隔开

u_{1} = \frac{- 2 + 2 33}{2 ( - 8 )}, u_{2} = \frac{- 2 - 2 33}{2 ( - 8 )}

u = \frac{- 2 + 2 33}{2 ( - 8 )} : - \frac{- 1 + 33}{8}

\frac{- 2 + 2 33}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 2 + 2 33}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 2 + 2 33}{- 16}

使用分式法则:

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

= - \frac{- 2 + 2 33}{16}

消掉

\frac{- 2 + 2 33}{16} : \frac{33 - 1}{8}

\frac{- 2 + 2 33}{16}

分解

- 2 + 233 : 2 (- 1 + 33)

- 2 + 233

改写为

= - 2 \cdot 1 + 233

因式分解出通项

2

= 2 (- 1 + 33)

= \frac{2 ( - 1 + 33 )}{16}

约分:

2

= \frac{- 1 + 33}{8}

= - \frac{33 - 1}{8}

= - \frac{- 1 + 33}{8}

u = \frac{- 2 - 2 33}{2 ( - 8 )} : \frac{1 + 33}{8}

\frac{- 2 - 2 33}{2 ( - 8 )}

去除括号:

(- a) = - a

= \frac{- 2 - 2 33}{- 2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 2 - 2 33}{- 16}

使用分式法则:

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

- 2 - 233 = - (2 + 233)

= \frac{2 + 2 33}{16}

分解

2 + 233 : 2 (1 + 33)

2 + 233

改写为

= 2 \cdot 1 + 233

因式分解出通项

2

= 2 (1 + 33)

= \frac{2 ( 1 + 33 )}{16}

约分:

2

= \frac{1 + 33}{8}

二次方程组的解是:

u = - \frac{- 1 + 33}{8}, u = \frac{1 + 33}{8}

u = sin (x)

代回

sin (x) = - \frac{- 1 + 33}{8}, sin (x) = \frac{1 + 33}{8}

sin (x) = - \frac{- 1 + 33}{8}, sin (x) = \frac{1 + 33}{8}

sin (x) = - \frac{- 1 + 33}{8} : x = arcsin (- \frac{- 1 + 33}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 33}{8}) + 2 πn

sin (x) = - \frac{- 1 + 33}{8}

使用反三角函数性质

sin (x) = - \frac{- 1 + 33}{8}

sin (x) = - \frac{- 1 + 33}{8}

的通解

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

x = arcsin (- \frac{- 1 + 33}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 33}{8}) + 2 πn

x = arcsin (- \frac{- 1 + 33}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 33}{8}) + 2 πn

sin (x) = \frac{1 + 33}{8} : x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

sin (x) = \frac{1 + 33}{8}

使用反三角函数性质

sin (x) = \frac{1 + 33}{8}

sin (x) = \frac{1 + 33}{8}

的通解

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

x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

合并所有解

x = arcsin (- \frac{- 1 + 33}{8}) + 2 πn, x = π + arcsin (\frac{- 1 + 33}{8}) + 2 πn, x = arcsin (\frac{1 + 33}{8}) + 2 πn, x = π - arcsin (\frac{1 + 33}{8}) + 2 πn

以小数形式表示解

x = - 0.63486 \dots + 2 πn, x = π + 0.63486 \dots + 2 πn, x = 1.00296 \dots + 2 πn, x = π - 1.00296 \dots + 2 πn

作图

流行的例子

tan^2(θ)+2tan(θ)=0,tan(2θ)+2tan(θ)=0

tan^{2} (θ) + 2 tan (θ) = 0, tan (2 θ) + 2 tan (θ) = 0

2cot(x)csc(x)-csc^2(x)=0

2 cot (x) csc (x) - csc^{2} (x) = 0

8tan(a)+3=2tan(a)+3

8 tan (a) + 3 = 2 tan (a) + 3

tan(θ/2)=(sqrt(3))/3

tan (\frac{θ}{2}) = \frac{3}{3}

solvefor x,cos(x)+1=0

so l v e f or x, cos (x) + 1 = 0