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受欢迎的三角函数 >

5cos(2θ)=-15cos(θ)-10

解答

5 cos (2 θ) = - 15 cos (θ) - 10

解答

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

+1

度数

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

求解步骤

5 cos (2 θ) = - 15 cos (θ) - 10

两边减去

- 15 cos (θ) - 10

5 cos (2 θ) + 15 cos (θ) + 10 = 0

使用三角恒等式改写

10 + 15 cos (θ) + 5 cos (2 θ)

使用倍角公式:

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

= 10 + 15 cos (θ) + 5 (2 cos^{2} (θ) - 1)

化简

10 + 15 cos (θ) + 5 (2 cos^{2} (θ) - 1) : 10 cos^{2} (θ) + 15 cos (θ) + 5

10 + 15 cos (θ) + 5 (2 cos^{2} (θ) - 1)

乘开

5 (2 cos^{2} (θ) - 1) : 10 cos^{2} (θ) - 5

5 (2 cos^{2} (θ) - 1)

使用分配律:

a (b - c) = ab - a c

a = 5, b = 2 cos^{2} (θ), c = 1

= 5 \cdot 2 cos^{2} (θ) - 5 \cdot 1

化简

5 \cdot 2 cos^{2} (θ) - 5 \cdot 1 : 10 cos^{2} (θ) - 5

5 \cdot 2 cos^{2} (θ) - 5 \cdot 1

数字相乘:

5 \cdot 2 = 10

= 10 cos^{2} (θ) - 5 \cdot 1

数字相乘:

5 \cdot 1 = 5

= 10 cos^{2} (θ) - 5

= 10 cos^{2} (θ) - 5

= 10 + 15 cos (θ) + 10 cos^{2} (θ) - 5

化简

10 + 15 cos (θ) + 10 cos^{2} (θ) - 5 : 10 cos^{2} (θ) + 15 cos (θ) + 5

10 + 15 cos (θ) + 10 cos^{2} (θ) - 5

对同类项分组

= 15 cos (θ) + 10 cos^{2} (θ) + 10 - 5

数字相加/相减:

10 - 5 = 5

= 10 cos^{2} (θ) + 15 cos (θ) + 5

= 10 cos^{2} (θ) + 15 cos (θ) + 5

= 10 cos^{2} (θ) + 15 cos (θ) + 5

5 + 10 cos^{2} (θ) + 15 cos (θ) = 0

用替代法求解

5 + 10 cos^{2} (θ) + 15 cos (θ) = 0

令

: cos (θ) = u

5 + 10 u^{2} + 15 u = 0

5 + 10 u^{2} + 15 u = 0 : u = - \frac{1}{2}, u = - 1

5 + 10 u^{2} + 15 u = 0

改写成标准形式

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

10 u^{2} + 15 u + 5 = 0

使用求根公式求解

10 u^{2} + 15 u + 5 = 0

二次方程求根公式:

若

a = 10, b = 15, c = 5

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 10 \cdot 5}{2 \cdot 10}

1 5^{2} - 4 \cdot 10 \cdot 5 = 5

1 5^{2} - 4 \cdot 10 \cdot 5

数字相乘:

4 \cdot 10 \cdot 5 = 200

= 1 5^{2} - 200

1 5^{2} = 225

= 225 - 200

数字相减:

225 - 200 = 25

= 25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 15 \pm 5}{2 \cdot 10}

将解分隔开

u_{1} = \frac{- 15 + 5}{2 \cdot 10}, u_{2} = \frac{- 15 - 5}{2 \cdot 10}

u = \frac{- 15 + 5}{2 \cdot 10} : - \frac{1}{2}

\frac{- 15 + 5}{2 \cdot 10}

数字相加/相减:

- 15 + 5 = - 10

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

数字相乘:

2 \cdot 10 = 20

= \frac{- 10}{20}

使用分式法则:

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

= - \frac{10}{20}

约分:

10

= - \frac{1}{2}

u = \frac{- 15 - 5}{2 \cdot 10} : - 1

\frac{- 15 - 5}{2 \cdot 10}

数字相减:

- 15 - 5 = - 20

= \frac{- 20}{2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 20}{20}

使用分式法则:

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

= - \frac{20}{20}

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 (θ)

代回

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

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

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

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

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

合并所有解

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

作图

流行的例子

2cos(x)-5=3sec(x)

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

1 = tan (c)

sin(a)=(sqrt(3))/2

sin (a) = \frac{3}{2}

tan^{2} (x) = 4

sec (x) = 3