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受欢迎的三角函数 >

6-sin(θ)=cos(2θ)

解答

6 - sin (θ) = cos (2 θ)

解答

θ \in R 无解

求解步骤

6 - sin (θ) = cos (2 θ)

两边减去

cos (2 θ)

6 - sin (θ) - cos (2 θ) = 0

使用三角恒等式改写

6 - cos (2 θ) - sin (θ)

使用倍角公式:

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

= 6 - (1 - 2 sin^{2} (θ)) - sin (θ)

化简

6 - (1 - 2 sin^{2} (θ)) - sin (θ) : 2 sin^{2} (θ) - sin (θ) + 5

6 - (1 - 2 sin^{2} (θ)) - sin (θ)

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

- (1 - 2 sin^{2} (θ))

打开括号

= - (1) - (- 2 sin^{2} (θ))

使用加减运算法则

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

= - 1 + 2 sin^{2} (θ)

= 6 - 1 + 2 sin^{2} (θ) - sin (θ)

数字相减:

6 - 1 = 5

= 2 sin^{2} (θ) - sin (θ) + 5

= 2 sin^{2} (θ) - sin (θ) + 5

5 - sin (θ) + 2 sin^{2} (θ) = 0

用替代法求解

5 - sin (θ) + 2 sin^{2} (θ) = 0

令

: sin (θ) = u

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

5 - u + 2 u^{2} = 0 : u = \frac{1}{4} + i \frac{39}{4}, u = \frac{1}{4} - i \frac{39}{4}

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

改写成标准形式

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

2 u^{2} - u + 5 = 0

使用求根公式求解

2 u^{2} - u + 5 = 0

二次方程求根公式:

若

a = 2, b = - 1, c = 5

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

化简

(- 1)^{2} - 4 \cdot 2 \cdot 5 : 39 i

(- 1)^{2} - 4 \cdot 2 \cdot 5

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 2 \cdot 5 = 40

4 \cdot 2 \cdot 5

数字相乘:

4 \cdot 2 \cdot 5 = 40

= 40

= 1 - 40

数字相减:

1 - 40 = - 39

= - 39

使用根式运算法则:

- a = - 1 a

- 39 = - 1 39

= - 1 39

使用虚数运算法则:

- 1 = i

= 39 i

u_{1, 2} = \frac{- ( - 1 ) \pm 39 i}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 39 i}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 39 i}{2 \cdot 2}

u = \frac{- ( - 1 ) + 39 i}{2 \cdot 2} : \frac{1}{4} + i \frac{39}{4}

\frac{- ( - 1 ) + 39 i}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 + 39 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 + 39 i}{4}

将

\frac{1 + 39 i}{4}

改写成标准复数形式:

\frac{1}{4} + \frac{39}{4} i

\frac{1 + 39 i}{4}

使用分式法则:

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

\frac{1 + 39 i}{4} = \frac{1}{4} + \frac{39 i}{4}

= \frac{1}{4} + \frac{39 i}{4}

= \frac{1}{4} + \frac{39}{4} i

u = \frac{- ( - 1 ) - 39 i}{2 \cdot 2} : \frac{1}{4} - i \frac{39}{4}

\frac{- ( - 1 ) - 39 i}{2 \cdot 2}

使用法则

- (- a) = a

= \frac{1 - 39 i}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{1 - 39 i}{4}

将

\frac{1 - 39 i}{4}

改写成标准复数形式:

\frac{1}{4} - \frac{39}{4} i

\frac{1 - 39 i}{4}

使用分式法则:

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

\frac{1 - 39 i}{4} = \frac{1}{4} - \frac{39 i}{4}

= \frac{1}{4} - \frac{39 i}{4}

= \frac{1}{4} - \frac{39}{4} i

二次方程组的解是:

u = \frac{1}{4} + i \frac{39}{4}, u = \frac{1}{4} - i \frac{39}{4}

u = sin (θ)

代回

sin (θ) = \frac{1}{4} + i \frac{39}{4}, sin (θ) = \frac{1}{4} - i \frac{39}{4}

sin (θ) = \frac{1}{4} + i \frac{39}{4}, sin (θ) = \frac{1}{4} - i \frac{39}{4}

sin (θ) = \frac{1}{4} + i \frac{39}{4} :

无解

sin (θ) = \frac{1}{4} + i \frac{39}{4}

无解

sin (θ) = \frac{1}{4} - i \frac{39}{4} :

无解

sin (θ) = \frac{1}{4} - i \frac{39}{4}

无解

合并所有解

θ \in R 无解

作图

流行的例子

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3tan^2(x)+5tan(x)+2=0

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