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受欢迎的三角函数 >

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

解答

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

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

两边减去

sin^{2} (x)

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

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

打开括号

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

使用加减运算法则

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

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

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

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

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

使用法则

- (- a) = a

= 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 \cdot 1}

将解分隔开

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

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

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

数字相乘:

2 \cdot 1 = 2

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

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

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

数字相乘:

2 \cdot 1 = 2

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

二次方程组的解是:

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

u = cos (x)

代回

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

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

无解

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

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

流行的例子

sin(θ)=(5pi)/3

sin (θ) = \frac{5 π}{3}

10 cos (θ) = 8

cos(θ)=(-18)/(sqrt(20)*\sqrt{26)}

cos (θ) = \frac{- 18}{20 \cdot 26}

sin(2x)=-cos(x),0<= x<= 2pi

sin (2 x) = - cos (x), 0 \leq x \leq 2 π

sin(θ)=(5pi)/6

sin (θ) = \frac{5 π}{6}