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受欢迎的三角函数 >

2cos^2(t)+7cos(t)+5=0

解答

2 cos^{2} (t) + 7 cos (t) + 5 = 0

解答

t = π + 2 πn

+1

度数

t = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

2 cos^{2} (t) + 7 cos (t) + 5 = 0

用替代法求解

2 cos^{2} (t) + 7 cos (t) + 5 = 0

令

: cos (t) = u

2 u^{2} + 7 u + 5 = 0

2 u^{2} + 7 u + 5 = 0 : u = - 1, u = - \frac{5}{2}

2 u^{2} + 7 u + 5 = 0

使用求根公式求解

2 u^{2} + 7 u + 5 = 0

二次方程求根公式:

若

a = 2, b = 7, c = 5

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 5 = 3

7^{2} - 4 \cdot 2 \cdot 5

数字相乘:

4 \cdot 2 \cdot 5 = 40

= 7^{2} - 40

7^{2} = 49

= 49 - 40

数字相减:

49 - 40 = 9

= 9

因式分解数字:

9 = 3^{2}

= 3^{2}

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 7 \pm 3}{2 \cdot 2}

将解分隔开

u_{1} = \frac{- 7 + 3}{2 \cdot 2}, u_{2} = \frac{- 7 - 3}{2 \cdot 2}

u = \frac{- 7 + 3}{2 \cdot 2} : - 1

\frac{- 7 + 3}{2 \cdot 2}

数字相加/相减:

- 7 + 3 = - 4

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

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

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 7 - 3}{2 \cdot 2} : - \frac{5}{2}

\frac{- 7 - 3}{2 \cdot 2}

数字相减:

- 7 - 3 = - 10

= \frac{- 10}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 10}{4}

使用分式法则:

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

= - \frac{10}{4}

约分:

2

= - \frac{5}{2}

二次方程组的解是:

u = - 1, u = - \frac{5}{2}

u = cos (t)

代回

cos (t) = - 1, cos (t) = - \frac{5}{2}

cos (t) = - 1, cos (t) = - \frac{5}{2}

cos (t) = - 1 : t = π + 2 πn

cos (t) = - 1

cos (t) = - 1

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}

t = π + 2 πn

t = π + 2 πn

cos (t) = - \frac{5}{2} :

无解

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

- 1 \leq cos (x) \leq 1

无解

合并所有解

t = π + 2 πn

作图

流行的例子

\frac{π}{6} = cos (2 t)

12cos^2(x)-cos(x)-1=0

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

solvefor t,x=acos^2(t)+bsin^2(t)

so l v e f or t, x = a cos^{2} (t) + b sin^{2} (t)

cos(3x)-sin(3x)=0

cos (3 x) - sin (3 x) = 0

8sin^2(x)cos^2(x)=1

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