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受欢迎的三角函数 >

sin^2(x)=3cos(x)

解答

sin^{2} (x) = 3 cos (x)

解答

x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

+1

度数

x = 72.37560 \dots^{\circ} + 36 0^{\circ} n, x = 287.62439 \dots^{\circ} + 36 0^{\circ} n

求解步骤

sin^{2} (x) = 3 cos (x)

两边减去

3 cos (x)

sin^{2} (x) - 3 cos (x) = 0

使用三角恒等式改写

sin^{2} (x) - 3 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) - 3 cos (x)

1 - cos^{2} (x) - 3 cos (x) = 0

用替代法求解

1 - cos^{2} (x) - 3 cos (x) = 0

令

: cos (x) = u

1 - u^{2} - 3 u = 0

1 - u^{2} - 3 u = 0 : u = - \frac{3 + 13}{2}, u = \frac{13 - 3}{2}

1 - u^{2} - 3 u = 0

改写成标准形式

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

- u^{2} - 3 u + 1 = 0

使用求根公式求解

- u^{2} - 3 u + 1 = 0

二次方程求根公式:

若

a = - 1, b = - 3, c = 1

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 3)^{2} - 4 (- 1) \cdot 1 = 13

(- 3)^{2} - 4 (- 1) \cdot 1

使用法则

- (- a) = a

= (- 3)^{2} + 4 \cdot 1 \cdot 1

使用指数法则:

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

若

n

是偶数

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 3^{2} + 4

3^{2} = 9

= 9 + 4

数字相加:

9 + 4 = 13

= 13

u_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 3 ) + 13}{2 ( - 1 )}, u_{2} = \frac{- ( - 3 ) - 13}{2 ( - 1 )}

u = \frac{- ( - 3 ) + 13}{2 ( - 1 )} : - \frac{3 + 13}{2}

\frac{- ( - 3 ) + 13}{2 ( - 1 )}

去除括号:

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

= \frac{3 + 13}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{3 + 13}{- 2}

使用分式法则:

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

= - \frac{3 + 13}{2}

u = \frac{- ( - 3 ) - 13}{2 ( - 1 )} : \frac{13 - 3}{2}

\frac{- ( - 3 ) - 13}{2 ( - 1 )}

去除括号:

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

= \frac{3 - 13}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{3 - 13}{- 2}

使用分式法则:

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

3 - 13 = - (13 - 3)

= \frac{13 - 3}{2}

二次方程组的解是:

u = - \frac{3 + 13}{2}, u = \frac{13 - 3}{2}

u = cos (x)

代回

cos (x) = - \frac{3 + 13}{2}, cos (x) = \frac{13 - 3}{2}

cos (x) = - \frac{3 + 13}{2}, cos (x) = \frac{13 - 3}{2}

cos (x) = - \frac{3 + 13}{2} :

无解

cos (x) = - \frac{3 + 13}{2}

- 1 \leq cos (x) \leq 1

无解

cos (x) = \frac{13 - 3}{2} : x = arccos (\frac{13 - 3}{2}) + 2 πn, x = 2 π - arccos (\frac{13 - 3}{2}) + 2 πn

cos (x) = \frac{13 - 3}{2}

使用反三角函数性质

cos (x) = \frac{13 - 3}{2}

cos (x) = \frac{13 - 3}{2}

的通解

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

x = arccos (\frac{13 - 3}{2}) + 2 πn, x = 2 π - arccos (\frac{13 - 3}{2}) + 2 πn

x = arccos (\frac{13 - 3}{2}) + 2 πn, x = 2 π - arccos (\frac{13 - 3}{2}) + 2 πn

合并所有解

x = arccos (\frac{13 - 3}{2}) + 2 πn, x = 2 π - arccos (\frac{13 - 3}{2}) + 2 πn

以小数形式表示解

x = 1.26319 \dots + 2 πn, x = 2 π - 1.26319 \dots + 2 πn

作图

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