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受欢迎的三角函数 >

cos(2x)+3cos(x)+2=0

解答

cos (2 x) + 3 cos (x) + 2 = 0

解答

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

+1

度数

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

cos (2 x) + 3 cos (x) + 2 = 0

使用三角恒等式改写

2 + cos (2 x) + 3 cos (x)

使用倍角公式:

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

= 2 + 2 cos^{2} (x) - 1 + 3 cos (x)

化简

2 + 2 cos^{2} (x) - 1 + 3 cos (x) : 2 cos^{2} (x) + 3 cos (x) + 1

2 + 2 cos^{2} (x) - 1 + 3 cos (x)

对同类项分组

= 2 cos^{2} (x) + 3 cos (x) + 2 - 1

数字相加/相减:

2 - 1 = 1

= 2 cos^{2} (x) + 3 cos (x) + 1

= 2 cos^{2} (x) + 3 cos (x) + 1

1 + 2 cos^{2} (x) + 3 cos (x) = 0

用替代法求解

1 + 2 cos^{2} (x) + 3 cos (x) = 0

令

: cos (x) = u

1 + 2 u^{2} + 3 u = 0

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

1 + 2 u^{2} + 3 u = 0

改写成标准形式

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

2 u^{2} + 3 u + 1 = 0

使用求根公式求解

2 u^{2} + 3 u + 1 = 0

二次方程求根公式:

若

a = 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

数字相乘:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数字相减:

9 - 8 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

\frac{- 3 + 1}{2 \cdot 2}

数字相加/相减:

- 3 + 1 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

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

\frac{- 3 - 1}{2 \cdot 2}

数字相减:

- 3 - 1 = - 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

二次方程组的解是:

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

u = cos (x)

代回

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

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

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

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

cos (x) = - \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}

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

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

合并所有解

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

作图

流行的例子

tan(θ)+1=sec(θ)

tan (θ) + 1 = sec (θ)

2sin(θ)=sqrt(2)

2 sin (θ) = 2

5cos(x)-sqrt(3)=3cos(x)

5 cos (x) - 3 = 3 cos (x)

tan (x) + 3 = 0

sqrt(2)sin(x)-1=0

2 sin (x) - 1 = 0