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受欢迎的三角函数 >

8cos^2(θ)+14cos(θ)-11=8cos(θ)-6

解答

8 cos^{2} (θ) + 14 cos (θ) - 11 = 8 cos (θ) - 6

解答

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

8 cos^{2} (θ) + 14 cos (θ) - 11 = 8 cos (θ) - 6

用替代法求解

8 cos^{2} (θ) + 14 cos (θ) - 11 = 8 cos (θ) - 6

令

: cos (θ) = u

8 u^{2} + 14 u - 11 = 8 u - 6

8 u^{2} + 14 u - 11 = 8 u - 6 : u = \frac{1}{2}, u = - \frac{5}{4}

8 u^{2} + 14 u - 11 = 8 u - 6

将

6

para o lado esquerdo

8 u^{2} + 14 u - 11 = 8 u - 6

两边加上

6

8 u^{2} + 14 u - 11 + 6 = 8 u - 6 + 6

化简

8 u^{2} + 14 u - 5 = 8 u

8 u^{2} + 14 u - 5 = 8 u

将

8 u

para o lado esquerdo

8 u^{2} + 14 u - 5 = 8 u

两边减去

8 u

8 u^{2} + 14 u - 5 - 8 u = 8 u - 8 u

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 8, b = 6, c = - 5

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

u_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 8 ( - 5 )}{2 \cdot 8}

6^{2} - 4 \cdot 8 (- 5) = 14

6^{2} - 4 \cdot 8 (- 5)

使用法则

- (- a) = a

= 6^{2} + 4 \cdot 8 \cdot 5

数字相乘:

4 \cdot 8 \cdot 5 = 160

= 6^{2} + 160

6^{2} = 36

= 36 + 160

数字相加:

36 + 160 = 196

= 196

因式分解数字:

196 = 1 4^{2}

= 1 4^{2}

使用根式运算法则:

n a^{n} = a

1 4^{2} = 14

= 14

u_{1, 2} = \frac{- 6 \pm 14}{2 \cdot 8}

将解分隔开

u_{1} = \frac{- 6 + 14}{2 \cdot 8}, u_{2} = \frac{- 6 - 14}{2 \cdot 8}

u = \frac{- 6 + 14}{2 \cdot 8} : \frac{1}{2}

\frac{- 6 + 14}{2 \cdot 8}

数字相加/相减:

- 6 + 14 = 8

= \frac{8}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

u = \frac{- 6 - 14}{2 \cdot 8} : - \frac{5}{4}

\frac{- 6 - 14}{2 \cdot 8}

数字相减:

- 6 - 14 = - 20

= \frac{- 20}{2 \cdot 8}

数字相乘:

2 \cdot 8 = 16

= \frac{- 20}{16}

使用分式法则:

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

= - \frac{20}{16}

约分:

4

= - \frac{5}{4}

二次方程组的解是:

u = \frac{1}{2}, u = - \frac{5}{4}

u = cos (θ)

代回

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{5}{4}

cos (θ) = \frac{1}{2}, cos (θ) = - \frac{5}{4}

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

cos (θ) = \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}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = - \frac{5}{4} :

无解

cos (θ) = - \frac{5}{4}

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

作图

流行的例子

sin^2(x)(tan^2(x))=1-sin^2(x)

sin^{2} (x) (tan^{2} (x)) = 1 - sin^{2} (x)

2sin^2(x)=3-5cos(x)

2 sin^{2} (x) = 3 - 5 cos (x)

-3cos(2θ)-18sin(θ)-8=-5sin(θ)-6

- 3 cos (2 θ) - 18 sin (θ) - 8 = - 5 sin (θ) - 6

tan^2(x)+4tan(x)=3

tan^{2} (x) + 4 tan (x) = 3

sin (1 - 2 x) = 1