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受欢迎的三角函数 >

cos^2(x)-5cos(x)+6=0

解答

cos^{2} (x) - 5 cos (x) + 6 = 0

解答

x \in R 无解

求解步骤

cos^{2} (x) - 5 cos (x) + 6 = 0

用替代法求解

cos^{2} (x) - 5 cos (x) + 6 = 0

令

: cos (x) = u

u^{2} - 5 u + 6 = 0

u^{2} - 5 u + 6 = 0 : u = 3, u = 2

u^{2} - 5 u + 6 = 0

使用求根公式求解

u^{2} - 5 u + 6 = 0

二次方程求根公式:

若

a = 1, b = - 5, c = 6

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

(- 5)^{2} - 4 \cdot 1 \cdot 6 = 1

(- 5)^{2} - 4 \cdot 1 \cdot 6

使用指数法则:

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

若

n

是偶数

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 1 \cdot 6

数字相乘:

4 \cdot 1 \cdot 6 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

u = \frac{- ( - 5 ) + 1}{2 \cdot 1} : 3

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

使用法则

- (- a) = a

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

数字相加:

5 + 1 = 6

= \frac{6}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

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

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

使用法则

- (- a) = a

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

数字相减:

5 - 1 = 4

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

数字相乘:

2 \cdot 1 = 2

= \frac{4}{2}

数字相除:

\frac{4}{2} = 2

= 2

二次方程组的解是:

u = 3, u = 2

u = cos (x)

代回

cos (x) = 3, cos (x) = 2

cos (x) = 3, cos (x) = 2

cos (x) = 3 :

无解

cos (x) = 3

- 1 \leq cos (x) \leq 1

无解

cos (x) = 2 :

无解

cos (x) = 2

- 1 \leq cos (x) \leq 1

无解

合并所有解

x \in R 无解

作图

流行的例子

cot^2(x/2)=3,0,0<= x<= 360

cot^{2} (\frac{x}{2}) = 3, 0, 0 \leq x \leq 360

sin(x)=(123sin(70))/(184.5)

sin (x) = \frac{123 sin ( 7 0 ^{\circ} )}{184.5}

cot^2(θ)=cot(θ),0<= θ<= 2pi

cot^{2} (θ) = cot (θ), 0 \leq θ \leq 2 π

cos (x) = \frac{9}{19}

cos (2 θ) = \frac{1}{4}