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受欢迎的三角函数 >

5-sin(θ)=cos(2θ)

解答

5 - sin (θ) = cos (2 θ)

解答

θ \in R 无解

求解步骤

5 - sin (θ) = cos (2 θ)

两边减去

cos (2 θ)

5 - sin (θ) - cos (2 θ) = 0

使用三角恒等式改写

5 - cos (2 θ) - sin (θ)

使用倍角公式:

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

= 5 - (1 - 2 sin^{2} (θ)) - sin (θ)

化简

5 - (1 - 2 sin^{2} (θ)) - sin (θ) : 2 sin^{2} (θ) - sin (θ) + 4

5 - (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)

= 5 - 1 + 2 sin^{2} (θ) - sin (θ)

数字相减:

5 - 1 = 4

= 2 sin^{2} (θ) - sin (θ) + 4

= 2 sin^{2} (θ) - sin (θ) + 4

4 - sin (θ) + 2 sin^{2} (θ) = 0

用替代法求解

4 - sin (θ) + 2 sin^{2} (θ) = 0

令

: sin (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

化简

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

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

(- 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 4 = 32

4 \cdot 2 \cdot 4

数字相乘:

4 \cdot 2 \cdot 4 = 32

= 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 \cdot 2}

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

= \frac{1 + 31 i}{4}

将

\frac{1 + 31 i}{4}

改写成标准复数形式:

\frac{1}{4} + \frac{31}{4} i

\frac{1 + 31 i}{4}

使用分式法则:

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

\frac{1 + 31 i}{4} = \frac{1}{4} + \frac{31 i}{4}

= \frac{1}{4} + \frac{31 i}{4}

= \frac{1}{4} + \frac{31}{4} i

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

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

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 2 = 4

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

将

\frac{1 - 31 i}{4}

改写成标准复数形式:

\frac{1}{4} - \frac{31}{4} i

\frac{1 - 31 i}{4}

使用分式法则:

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

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

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

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

二次方程组的解是:

u = \frac{1}{4} + i \frac{31}{4}, u = \frac{1}{4} - i \frac{31}{4}

u = sin (θ)

代回

sin (θ) = \frac{1}{4} + i \frac{31}{4}, sin (θ) = \frac{1}{4} - i \frac{31}{4}

sin (θ) = \frac{1}{4} + i \frac{31}{4}, sin (θ) = \frac{1}{4} - i \frac{31}{4}

sin (θ) = \frac{1}{4} + i \frac{31}{4} :

无解

sin (θ) = \frac{1}{4} + i \frac{31}{4}

无解

sin (θ) = \frac{1}{4} - i \frac{31}{4} :

无解

sin (θ) = \frac{1}{4} - i \frac{31}{4}

无解

合并所有解

θ \in R 无解

作图

流行的例子

8sin(θ)+4sqrt(3)=0

8 sin (θ) + 43 = 0

solvefor x,sin(x)=(sqrt(3))/2

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tan (α) = 2

sin(2x+pi/4)=(sqrt(2))/2

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