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 >

7cos(θ)=sqrt(2)+9cos(θ)

解答

7 cos (θ) = 2 + 9 cos (θ)

解答

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

+1

度数

θ = 13 5^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n

求解步骤

7 cos (θ) = 2 + 9 cos (θ)

用替代法求解

7 cos (θ) = 2 + 9 cos (θ)

令

: cos (θ) = u

7 u = 2 + 9 u

7 u = 2 + 9 u : u = - \frac{2}{2}

7 u = 2 + 9 u

将

9 u

para o lado esquerdo

7 u = 2 + 9 u

两边减去

9 u

7 u - 9 u = 2 + 9 u - 9 u

化简

- 2 u = 2

- 2 u = 2

两边除以

- 2

- 2 u = 2

两边除以

- 2

\frac{- 2 u}{- 2} = \frac{2}{- 2}

化简

u = - \frac{2}{2}

u = - \frac{2}{2}

u = cos (θ)

代回

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{2} : θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

cos (θ) = - \frac{2}{2}

cos (θ) = - \frac{2}{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{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

合并所有解

θ = \frac{3 π}{4} + 2 πn, θ = \frac{5 π}{4} + 2 πn

作图

流行的例子

sec(x)*cot(x)-2cot(x)+sec(x)=2

3sin^2(x)-10sin(x)-8=0

cos(7x)=sin(4x+2)

12cos(β)-5sin(β)=4.7

cos(x)=(sqrt(8))/3