zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

8cos^2(θ)+13=14cos(θ)+8

解答

8 cos^{2} (θ) + 13 = 14 cos (θ) + 8

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) + 13 = 14 cos (θ) + 8

用替代法求解

8 cos^{2} (θ) + 13 = 14 cos (θ) + 8

令

: cos (θ) = u

8 u^{2} + 13 = 14 u + 8

8 u^{2} + 13 = 14 u + 8 : u = \frac{5}{4}, u = \frac{1}{2}

8 u^{2} + 13 = 14 u + 8

将

8

para o lado esquerdo

8 u^{2} + 13 = 14 u + 8

两边减去

8

8 u^{2} + 13 - 8 = 14 u + 8 - 8

化简

8 u^{2} + 5 = 14 u

8 u^{2} + 5 = 14 u

将

14 u

para o lado esquerdo

8 u^{2} + 5 = 14 u

两边减去

14 u

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

化简

8 u^{2} + 5 - 14 u = 0

8 u^{2} + 5 - 14 u = 0

改写成标准形式

a x^{2} + b x + c = 0

8 u^{2} - 14 u + 5 = 0

使用求根公式求解

8 u^{2} - 14 u + 5 = 0

二次方程求根公式:

若

a = 8, b = - 14, c = 5

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 8 \cdot 5}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 8 \cdot 5}{2 \cdot 8}

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

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

使用指数法则:

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

若

n

是偶数

(- 14)^{2} = 1 4^{2}

= 1 4^{2} - 4 \cdot 8 \cdot 5

数字相乘:

4 \cdot 8 \cdot 5 = 160

= 1 4^{2} - 160

1 4^{2} = 196

= 196 - 160

数字相减:

196 - 160 = 36

= 36

因式分解数字:

36 = 6^{2}

= 6^{2}

使用根式运算法则:

n a^{n} = a

6^{2} = 6

= 6

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

14 + 6 = 20

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

数字相乘:

2 \cdot 8 = 16

= \frac{20}{16}

约分:

4

= \frac{5}{4}

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

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

使用法则

- (- a) = a

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

数字相减:

14 - 6 = 8

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

数字相乘:

2 \cdot 8 = 16

= \frac{8}{16}

约分:

8

= \frac{1}{2}

二次方程组的解是:

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

u = cos (θ)

代回

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

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

cos (θ) = \frac{5}{4} :

无解

cos (θ) = \frac{5}{4}

- 1 \leq cos (x) \leq 1

无解

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

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

cos(3x)=-1/2 ,0<= x<2pi

cos (3 x) = - \frac{1}{2}, 0 \leq x < 2 π

cos (5 t) = 0

tan(4x)=(sqrt(3))/3

tan (4 x) = \frac{3}{3}

tan^2(x)=tan^2(x)sec(x)

tan^{2} (x) = tan^{2} (x) sec (x)

2tan^2(x)=sec^2(x)

2 tan^{2} (x) = sec^{2} (x)