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受欢迎的三角函数 >

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

解答

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

解答

x \in R 无解

求解步骤

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

使用三角恒等式改写

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

使用倍角公式:

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

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

化简

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

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

乘开

2 (1 - 2 sin^{2} (x)) : 2 - 4 sin^{2} (x)

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

使用分配律:

a (b - c) = ab - a c

a = 2, b = 1, c = 2 sin^{2} (x)

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

化简

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) : 2 - 4 sin^{2} (x)

2 \cdot 1 - 2 \cdot 2 sin^{2} (x)

数字相乘:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (x)

数字相乘:

2 \cdot 2 = 4

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

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

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

化简

- 4 + sin (x) + 2 - 4 sin^{2} (x) : sin (x) - 4 sin^{2} (x) - 2

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

对同类项分组

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

数字相加/相减:

- 4 + 2 = - 2

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

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

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

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

用替代法求解

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

令

: sin (x) = u

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

- 2 + u - 4 u^{2} = 0 : u = \frac{1}{8} - i \frac{31}{8}, u = \frac{1}{8} + i \frac{31}{8}

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 4, b = 1, c = - 2

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

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

化简

1^{2} - 4 (- 4) (- 2) : 31 i

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 4) (- 2)

使用法则

- (- a) = a

= 1 - 4 \cdot 4 \cdot 2

数字相乘:

4 \cdot 4 \cdot 2 = 32

= 1 - 32

数字相减:

1 - 32 = - 31

= - 31

使用根式运算法则:

- a = - 1 a

- 31 = - 1 31

= - 1 31

使用虚数运算法则:

- 1 = i

= 31 i

u_{1, 2} = \frac{- 1 \pm 31 i}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 1 + 31 i}{2 ( - 4 )}, u_{2} = \frac{- 1 - 31 i}{2 ( - 4 )}

u = \frac{- 1 + 31 i}{2 ( - 4 )} : \frac{1}{8} - i \frac{31}{8}

\frac{- 1 + 31 i}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 1 + 31 i}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 + 31 i}{- 8}

使用分式法则:

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

= - \frac{- 1 + 31 i}{8}

将

- \frac{- 1 + 31 i}{8}

改写成标准复数形式:

\frac{1}{8} - \frac{31}{8} i

- \frac{- 1 + 31 i}{8}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 + 31 i}{8} = - (- \frac{1}{8}) - (\frac{31 i}{8})

= - (- \frac{1}{8}) - (\frac{31 i}{8})

去除括号:

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

= \frac{1}{8} - \frac{31 i}{8}

= \frac{1}{8} - \frac{31}{8} i

u = \frac{- 1 - 31 i}{2 ( - 4 )} : \frac{1}{8} + i \frac{31}{8}

\frac{- 1 - 31 i}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 1 - 31 i}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 1 - 31 i}{- 8}

使用分式法则:

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

= - \frac{- 1 - 31 i}{8}

将

- \frac{- 1 - 31 i}{8}

改写成标准复数形式:

\frac{1}{8} + \frac{31}{8} i

- \frac{- 1 - 31 i}{8}

使用分式法则:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 1 - 31 i}{8} = - (- \frac{1}{8}) - (- \frac{31 i}{8})

= - (- \frac{1}{8}) - (- \frac{31 i}{8})

使用法则

- (- a) = a

= \frac{1}{8} + \frac{31 i}{8}

= \frac{1}{8} + \frac{31}{8} i

二次方程组的解是:

u = \frac{1}{8} - i \frac{31}{8}, u = \frac{1}{8} + i \frac{31}{8}

u = sin (x)

代回

sin (x) = \frac{1}{8} - i \frac{31}{8}, sin (x) = \frac{1}{8} + i \frac{31}{8}

sin (x) = \frac{1}{8} - i \frac{31}{8}, sin (x) = \frac{1}{8} + i \frac{31}{8}

sin (x) = \frac{1}{8} - i \frac{31}{8} :

无解

sin (x) = \frac{1}{8} - i \frac{31}{8}

无解

sin (x) = \frac{1}{8} + i \frac{31}{8} :

无解

sin (x) = \frac{1}{8} + i \frac{31}{8}

无解

合并所有解

x \in R 无解

作图

流行的例子

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(sin(115))/(20)=(sin(b))/(11)

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( b )}{11}

cos(θ)=(25)/(sqrt(714))

cos (θ) = \frac{25}{714}

3sec^2(x)-4=0,0<= x<2pi

3 sec^{2} (x) - 4 = 0, 0 \leq x < 2 π

3cos(2x)+7sin(x)-5=0

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