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受欢迎的三角函数 >

2sin(x)=(4cos(x)-1)/(tan(x))

解答

2 sin (x) = \frac{4 cos ( x ) - 1}{tan ( x )}

解答

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

+1

度数

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n

求解步骤

2 sin (x) = \frac{4 cos ( x ) - 1}{tan ( x )}

两边减去

\frac{4 cos ( x ) - 1}{tan ( x )}

2 sin (x) - \frac{4 cos ( x ) - 1}{tan ( x )} = 0

化简

2 sin (x) - \frac{4 cos ( x ) - 1}{tan ( x )} : \frac{2 sin ( x ) tan ( x ) - 4 cos ( x ) + 1}{tan ( x )}

2 sin (x) - \frac{4 cos ( x ) - 1}{tan ( x )}

将项转换为分式:

2 sin (x) = \frac{2 sin ( x ) tan ( x )}{tan ( x )}

= \frac{2 sin ( x ) tan ( x )}{tan ( x )} - \frac{4 cos ( x ) - 1}{tan ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 sin ( x ) tan ( x ) - ( 4 cos ( x ) - 1 )}{tan ( x )}

- (4 cos (x) - 1) : - 4 cos (x) + 1

- (4 cos (x) - 1)

打开括号

= - (4 cos (x)) - (- 1)

使用加减运算法则

- (- a) = a, - (a) = - a

= - 4 cos (x) + 1

= \frac{2 sin ( x ) tan ( x ) - 4 cos ( x ) + 1}{tan ( x )}

\frac{2 sin ( x ) tan ( x ) - 4 cos ( x ) + 1}{tan ( x )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin (x) tan (x) - 4 cos (x) + 1 = 0

使用三角恒等式改写

1 - 4 cos (x) + 2 sin (x) tan (x)

使用基本三角恒等式:

tan (x) = \frac{sin ( x )}{cos ( x )}

= 1 - 4 cos (x) + 2 sin (x) \frac{sin ( x )}{cos ( x )}

2 sin (x) \frac{sin ( x )}{cos ( x )} = \frac{2 sin ^{2} ( x )}{cos ( x )}

2 sin (x) \frac{sin ( x )}{cos ( x )}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( x ) \cdot 2 sin ( x )}{cos ( x )}

sin (x) \cdot 2 sin (x) = 2 sin^{2} (x)

sin (x) \cdot 2 sin (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

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

数字相加:

1 + 1 = 2

= 2 sin^{2} (x)

= \frac{2 sin ^{2} ( x )}{cos ( x )}

= 1 - 4 cos (x) + \frac{2 sin ^{2} ( x )}{cos ( x )}

使用毕达哥拉斯恒等式:

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

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

= 1 + \frac{2 ( 1 - cos ^{2} ( x ) )}{cos ( x )} - 4 cos (x)

合并分式

\frac{2 ( - cos ^{2} ( x ) + 1 )}{cos ( x )} - 4 cos (x) : \frac{2 - 6 cos ^{2} ( x )}{cos ( x )}

\frac{2 ( - cos ^{2} ( x ) + 1 )}{cos ( x )} - 4 cos (x)

将项转换为分式:

4 cos (x) = \frac{4 cos ( x ) cos ( x )}{cos ( x )}

= \frac{2 ( 1 - cos ^{2} ( x ) )}{cos ( x )} - \frac{4 cos ( x ) cos ( x )}{cos ( x )}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 ( 1 - cos ^{2} ( x ) ) - 4 cos ( x ) cos ( x )}{cos ( x )}

2 (1 - cos^{2} (x)) - 4 cos (x) cos (x) = 2 (1 - cos^{2} (x)) - 4 cos^{2} (x)

2 (1 - cos^{2} (x)) - 4 cos (x) cos (x)

4 cos (x) cos (x) = 4 cos^{2} (x)

4 cos (x) cos (x)

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

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

= 4 cos^{1 + 1} (x)

数字相加:

1 + 1 = 2

= 4 cos^{2} (x)

= 2 (- cos^{2} (x) + 1) - 4 cos^{2} (x)

= \frac{2 ( - cos ^{2} ( x ) + 1 ) - 4 cos ^{2} ( x )}{cos ( x )}

乘开

2 (1 - cos^{2} (x)) - 4 cos^{2} (x) : 2 - 6 cos^{2} (x)

2 (1 - cos^{2} (x)) - 4 cos^{2} (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

2 \cdot 1 = 2

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

= 2 - 2 cos^{2} (x) - 4 cos^{2} (x)

同类项相加:

- 2 cos^{2} (x) - 4 cos^{2} (x) = - 6 cos^{2} (x)

= 2 - 6 cos^{2} (x)

= \frac{2 - 6 cos ^{2} ( x )}{cos ( x )}

= \frac{2 - 6 cos ^{2} ( x )}{cos ( x )} + 1

1 + \frac{2 - 6 cos ^{2} ( x )}{cos ( x )} = 0

1 + \frac{2 - 6 cos ^{2} ( x )}{cos ( x )} = 0

用替代法求解

1 + \frac{2 - 6 cos ^{2} ( x )}{cos ( x )} = 0

令

: cos (x) = u

1 + \frac{2 - 6 u ^{2}}{u} = 0

1 + \frac{2 - 6 u ^{2}}{u} = 0 : u = - \frac{1}{2}, u = \frac{2}{3}

1 + \frac{2 - 6 u ^{2}}{u} = 0

在两边乘以

u

1 + \frac{2 - 6 u ^{2}}{u} = 0

在两边乘以

u

1 \cdot u + \frac{2 - 6 u ^{2}}{u} u = 0 \cdot u

化简

1 \cdot u + \frac{2 - 6 u ^{2}}{u} u = 0 \cdot u

化简

1 \cdot u : u

1 \cdot u

乘以:

1 \cdot u = u

= u

化简

\frac{2 - 6 u ^{2}}{u} u : 2 - 6 u^{2}

\frac{2 - 6 u ^{2}}{u} u

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 2 - 6 u ^{2} ) u}{u}

约分:

u

= 2 - 6 u^{2}

化简

0 \cdot u : 0

0 \cdot u

使用法则

0 \cdot a = 0

= 0

u + 2 - 6 u^{2} = 0

u + 2 - 6 u^{2} = 0

u + 2 - 6 u^{2} = 0

解

u + 2 - 6 u^{2} = 0 : u = - \frac{1}{2}, u = \frac{2}{3}

u + 2 - 6 u^{2} = 0

改写成标准形式

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

- 6 u^{2} + u + 2 = 0

使用求根公式求解

- 6 u^{2} + u + 2 = 0

二次方程求根公式:

若

a = - 6, b = 1, c = 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 2}{2 ( - 6 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 6 ) \cdot 2}{2 ( - 6 )}

1^{2} - 4 (- 6) \cdot 2 = 7

1^{2} - 4 (- 6) \cdot 2

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 6) \cdot 2

使用法则

- (- a) = a

= 1 + 4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 1 + 48

数字相加:

1 + 48 = 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{- 1 \pm 7}{2 ( - 6 )}

将解分隔开

u_{1} = \frac{- 1 + 7}{2 ( - 6 )}, u_{2} = \frac{- 1 - 7}{2 ( - 6 )}

u = \frac{- 1 + 7}{2 ( - 6 )} : - \frac{1}{2}

\frac{- 1 + 7}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 1 + 7}{- 2 \cdot 6}

数字相加/相减:

- 1 + 7 = 6

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

数字相乘:

2 \cdot 6 = 12

= \frac{6}{- 12}

使用分式法则:

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

= - \frac{6}{12}

约分:

6

= - \frac{1}{2}

u = \frac{- 1 - 7}{2 ( - 6 )} : \frac{2}{3}

\frac{- 1 - 7}{2 ( - 6 )}

去除括号:

(- a) = - a

= \frac{- 1 - 7}{- 2 \cdot 6}

数字相减:

- 1 - 7 = - 8

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

数字相乘:

2 \cdot 6 = 12

= \frac{- 8}{- 12}

使用分式法则:

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

= \frac{8}{12}

约分:

4

= \frac{2}{3}

二次方程组的解是:

u = - \frac{1}{2}, u = \frac{2}{3}

u = - \frac{1}{2}, u = \frac{2}{3}

验证解

找到无定义的点（奇点）:

u = 0

取

1 + \frac{2 - 6 u ^{2}}{u}

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u = - \frac{1}{2}, u = \frac{2}{3}

u = cos (x)

代回

cos (x) = - \frac{1}{2}, cos (x) = \frac{2}{3}

cos (x) = - \frac{1}{2}, cos (x) = \frac{2}{3}

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

cos (x) = \frac{2}{3}

使用反三角函数性质

cos (x) = \frac{2}{3}

cos (x) = \frac{2}{3}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

x = arccos (\frac{2}{3}) + 2 πn, x = 2 π - arccos (\frac{2}{3}) + 2 πn

合并所有解

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

以小数形式表示解

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn

作图

流行的例子

6 cos (5 x) = 3

2sin(2θ)cos(θ)+cos(2θ)sin(θ)=0

2 sin (2 θ) cos (θ) + cos (2 θ) sin (θ) = 0

2sin^2(x)=2+cos(x),0<= x<= 2pi

2 sin^{2} (x) = 2 + cos (x), 0 \leq x \leq 2 π

sec(x)+sqrt(2)=0

sec (x) + 2 = 0

sin^2(x)+cos(x)-1=0

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