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受欢迎的三角函数 >

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

解答

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

解答

x = 1.33247 \dots + 2 πn, x = 2 π - 1.33247 \dots + 2 πn

+1

度数

x = 76.34541 \dots^{\circ} + 36 0^{\circ} n, x = 283.65458 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

4 cos (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

= 1 - cos^{2} (x) - 4 cos (x)

1 - cos^{2} (x) - 4 cos (x) = 0

用替代法求解

1 - cos^{2} (x) - 4 cos (x) = 0

令

: cos (x) = u

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

1 - u^{2} - 4 u = 0 : u = - 2 - 5, u = 5 - 2

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

改写成标准形式

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

- u^{2} - 4 u + 1 = 0

使用求根公式求解

- u^{2} - 4 u + 1 = 0

二次方程求根公式:

若

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

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

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

(- 4)^{2} - 4 (- 1) \cdot 1 = 25

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

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

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4^{2} + 4

4^{2} = 16

= 16 + 4

数字相加:

16 + 4 = 20

= 20

20

质因数分解:

2^{2} \cdot 5

20

20

除以

220 = 10 \cdot 2

= 2 \cdot 10

10

除以

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

使用根式运算法则:

n ab = n a n b

= 5 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 25

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

将解分隔开

u_{1} = \frac{- ( - 4 ) + 2 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 4 ) - 2 5}{2 ( - 1 )}

u = \frac{- ( - 4 ) + 2 5}{2 ( - 1 )} : - 2 - 5

\frac{- ( - 4 ) + 2 5}{2 ( - 1 )}

去除括号:

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

= \frac{4 + 2 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 + 2 5}{- 2}

使用分式法则:

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

= - \frac{4 + 2 5}{2}

消掉

\frac{4 + 2 5}{2} : 2 + 5

\frac{4 + 2 5}{2}

分解

4 + 25 : 2 (2 + 5)

4 + 25

改写为

= 2 \cdot 2 + 25

因式分解出通项

2

= 2 (2 + 5)

= \frac{2 ( 2 + 5 )}{2}

数字相除:

\frac{2}{2} = 1

= 2 + 5

= - (2 + 5)

打开括号

= - (2) - (5)

使用加减运算法则

+ (- a) = - a

= - 2 - 5

u = \frac{- ( - 4 ) - 2 5}{2 ( - 1 )} : 5 - 2

\frac{- ( - 4 ) - 2 5}{2 ( - 1 )}

去除括号:

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

= \frac{4 - 2 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{4 - 2 5}{- 2}

使用分式法则:

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

4 - 25 = - (25 - 4)

= \frac{2 5 - 4}{2}

分解

25 - 4 : 2 (5 - 2)

25 - 4

改写为

= 25 - 2 \cdot 2

因式分解出通项

2

= 2 (5 - 2)

= \frac{2 ( 5 - 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 5 - 2

二次方程组的解是:

u = - 2 - 5, u = 5 - 2

u = cos (x)

代回

cos (x) = - 2 - 5, cos (x) = 5 - 2

cos (x) = - 2 - 5, cos (x) = 5 - 2

cos (x) = - 2 - 5 :

无解

cos (x) = - 2 - 5

- 1 \leq cos (x) \leq 1

无解

cos (x) = 5 - 2 : x = arccos (5 - 2) + 2 πn, x = 2 π - arccos (5 - 2) + 2 πn

cos (x) = 5 - 2

使用反三角函数性质

cos (x) = 5 - 2

cos (x) = 5 - 2

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (5 - 2) + 2 πn, x = 2 π - arccos (5 - 2) + 2 πn

x = arccos (5 - 2) + 2 πn, x = 2 π - arccos (5 - 2) + 2 πn

合并所有解

x = arccos (5 - 2) + 2 πn, x = 2 π - arccos (5 - 2) + 2 πn

以小数形式表示解

x = 1.33247 \dots + 2 πn, x = 2 π - 1.33247 \dots + 2 πn

作图

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