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受欢迎的三角函数 >

14+7cos(2x)=21cos(x)

解答

14 + 7 cos (2 x) = 21 cos (x)

解答

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

+1

度数

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

求解步骤

14 + 7 cos (2 x) = 21 cos (x)

两边减去

21 cos (x)

14 + 7 cos (2 x) - 21 cos (x) = 0

使用三角恒等式改写

14 - 21 cos (x) + 7 cos (2 x)

使用倍角公式:

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

= 14 - 21 cos (x) + 7 (2 cos^{2} (x) - 1)

化简

14 - 21 cos (x) + 7 (2 cos^{2} (x) - 1) : 14 cos^{2} (x) - 21 cos (x) + 7

14 - 21 cos (x) + 7 (2 cos^{2} (x) - 1)

乘开

7 (2 cos^{2} (x) - 1) : 14 cos^{2} (x) - 7

7 (2 cos^{2} (x) - 1)

使用分配律:

a (b - c) = ab - a c

a = 7, b = 2 cos^{2} (x), c = 1

= 7 \cdot 2 cos^{2} (x) - 7 \cdot 1

化简

7 \cdot 2 cos^{2} (x) - 7 \cdot 1 : 14 cos^{2} (x) - 7

7 \cdot 2 cos^{2} (x) - 7 \cdot 1

数字相乘:

7 \cdot 2 = 14

= 14 cos^{2} (x) - 7 \cdot 1

数字相乘:

7 \cdot 1 = 7

= 14 cos^{2} (x) - 7

= 14 cos^{2} (x) - 7

= 14 - 21 cos (x) + 14 cos^{2} (x) - 7

化简

14 - 21 cos (x) + 14 cos^{2} (x) - 7 : 14 cos^{2} (x) - 21 cos (x) + 7

14 - 21 cos (x) + 14 cos^{2} (x) - 7

对同类项分组

= - 21 cos (x) + 14 cos^{2} (x) + 14 - 7

数字相加/相减:

14 - 7 = 7

= 14 cos^{2} (x) - 21 cos (x) + 7

= 14 cos^{2} (x) - 21 cos (x) + 7

= 14 cos^{2} (x) - 21 cos (x) + 7

7 + 14 cos^{2} (x) - 21 cos (x) = 0

用替代法求解

7 + 14 cos^{2} (x) - 21 cos (x) = 0

令

: cos (x) = u

7 + 14 u^{2} - 21 u = 0

7 + 14 u^{2} - 21 u = 0 : u = 1, u = \frac{1}{2}

7 + 14 u^{2} - 21 u = 0

改写成标准形式

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

14 u^{2} - 21 u + 7 = 0

使用求根公式求解

14 u^{2} - 21 u + 7 = 0

二次方程求根公式:

若

a = 14, b = - 21, c = 7

u_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 14 \cdot 7}{2 \cdot 14}

u_{1, 2} = \frac{- ( - 21 ) \pm ( - 21 ) ^{2} - 4 \cdot 14 \cdot 7}{2 \cdot 14}

(- 21)^{2} - 4 \cdot 14 \cdot 7 = 7

(- 21)^{2} - 4 \cdot 14 \cdot 7

使用指数法则:

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

若

n

是偶数

(- 21)^{2} = 2 1^{2}

= 2 1^{2} - 4 \cdot 14 \cdot 7

数字相乘:

4 \cdot 14 \cdot 7 = 392

= 2 1^{2} - 392

2 1^{2} = 441

= 441 - 392

数字相减:

441 - 392 = 49

= 49

因式分解数字:

49 = 7^{2}

= 7^{2}

使用根式运算法则:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 21 ) \pm 7}{2 \cdot 14}

将解分隔开

u_{1} = \frac{- ( - 21 ) + 7}{2 \cdot 14}, u_{2} = \frac{- ( - 21 ) - 7}{2 \cdot 14}

u = \frac{- ( - 21 ) + 7}{2 \cdot 14} : 1

\frac{- ( - 21 ) + 7}{2 \cdot 14}

使用法则

- (- a) = a

= \frac{21 + 7}{2 \cdot 14}

数字相加:

21 + 7 = 28

= \frac{28}{2 \cdot 14}

数字相乘:

2 \cdot 14 = 28

= \frac{28}{28}

使用法则

\frac{a}{a} = 1

= 1

u = \frac{- ( - 21 ) - 7}{2 \cdot 14} : \frac{1}{2}

\frac{- ( - 21 ) - 7}{2 \cdot 14}

使用法则

- (- a) = a

= \frac{21 - 7}{2 \cdot 14}

数字相减:

21 - 7 = 14

= \frac{14}{2 \cdot 14}

数字相乘:

2 \cdot 14 = 28

= \frac{14}{28}

约分:

14

= \frac{1}{2}

二次方程组的解是:

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

u = cos (x)

代回

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

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

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

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

合并所有解

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

作图

流行的例子

sin(3x)=0,0<= x<2pi

sin (3 x) = 0, 0 \leq x < 2 π

sin(θ)=-1/5 ,pi<θ<(3pi)/2 ,cos(θ)

sin (θ) = - \frac{1}{5}, π < θ < \frac{3 π}{2}, cos (θ)

(sin(x))/(37)=(sin(30))/(30)

\frac{sin ( x )}{37} = \frac{sin ( 3 0 ^{\circ} )}{30}

solvefor t,x=arccos(1/(sqrt(1+t^2)))

so l v e f or t, x = arccos (\frac{1}{1 + t ^{2}})

(1-sin(x))(1+cos(x))=cos^2(x)

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