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受欢迎的三角函数 >

6(cos(b))^2+cos(b)-1=0

解答

6 (cos (b))^{2} + cos (b) - 1 = 0

解答

b = 1.23095 \dots + 2 πn, b = 2 π - 1.23095 \dots + 2 πn, b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

+1

度数

b = 70.52877 \dots^{\circ} + 36 0^{\circ} n, b = 289.47122 \dots^{\circ} + 36 0^{\circ} n, b = 12 0^{\circ} + 36 0^{\circ} n, b = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

6 (cos (b))^{2} + cos (b) - 1 = 0

用替代法求解

6 (cos (b))^{2} + cos (b) - 1 = 0

令

: cos (b) = u

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

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 6 (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 1

数字相乘:

4 \cdot 6 \cdot 1 = 24

= 1 + 24

数字相加:

1 + 24 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 6}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 6}, u_{2} = \frac{- 1 - 5}{2 \cdot 6}

u = \frac{- 1 + 5}{2 \cdot 6} : \frac{1}{3}

\frac{- 1 + 5}{2 \cdot 6}

数字相加/相减:

- 1 + 5 = 4

= \frac{4}{2 \cdot 6}

数字相乘:

2 \cdot 6 = 12

= \frac{4}{12}

约分:

4

= \frac{1}{3}

u = \frac{- 1 - 5}{2 \cdot 6} : - \frac{1}{2}

\frac{- 1 - 5}{2 \cdot 6}

数字相减:

- 1 - 5 = - 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 6}{12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

二次方程组的解是:

u = \frac{1}{3}, u = - \frac{1}{2}

u = cos (b)

代回

cos (b) = \frac{1}{3}, cos (b) = - \frac{1}{2}

cos (b) = \frac{1}{3}, cos (b) = - \frac{1}{2}

cos (b) = \frac{1}{3} : b = arccos (\frac{1}{3}) + 2 πn, b = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (b) = \frac{1}{3}

使用反三角函数性质

cos (b) = \frac{1}{3}

cos (b) = \frac{1}{3}

的通解

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

b = arccos (\frac{1}{3}) + 2 πn, b = 2 π - arccos (\frac{1}{3}) + 2 πn

b = arccos (\frac{1}{3}) + 2 πn, b = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (b) = - \frac{1}{2} : b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

cos (b) = - \frac{1}{2}

cos (b) = - \frac{1}{2}

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

合并所有解

b = arccos (\frac{1}{3}) + 2 πn, b = 2 π - arccos (\frac{1}{3}) + 2 πn, b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

以小数形式表示解

b = 1.23095 \dots + 2 πn, b = 2 π - 1.23095 \dots + 2 πn, b = \frac{2 π}{3} + 2 πn, b = \frac{4 π}{3} + 2 πn

作图

流行的例子

sin(θ)=(1.33)/(2.42)

sin (θ) = \frac{1.33}{2.42}

2tan(x)=3-tan^2(x),-2pi<= ,x<= pi

2 tan (x) = 3 - tan^{2} (x), - 2 π \leq, x \leq π

-25=(29cos(x))*3.7-4.9*3.7^2

- 25 = (29 cos (x)) \cdot 3.7 - 4.9 \cdot 3. 7^{2}

solvefor t,x=5(sin(t)+cos(t))

so l v e f or t, x = 5 (sin (t) + cos (t))

2/2*sqrt(3)*cos(x)-2/2*sin(x)=0

\frac{2}{2} \cdot 3 \cdot cos (x) - \frac{2}{2} \cdot sin (x) = 0