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受欢迎的三角函数 >

4cos(θ)-1=2sin(θ)tan(θ)

解答

4 cos (θ) - 1 = 2 sin (θ) tan (θ)

解答

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

+1

度数

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

求解步骤

4 cos (θ) - 1 = 2 sin (θ) tan (θ)

两边减去

2 sin (θ) tan (θ)

4 cos (θ) - 1 - 2 sin (θ) tan (θ) = 0

使用三角恒等式改写

- 1 + 4 cos (θ) - 2 sin (θ) tan (θ)

使用基本三角恒等式:

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

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

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

2 sin (θ) \frac{sin ( θ )}{cos ( θ )}

分式相乘:

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

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

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

sin (θ) \cdot 2 sin (θ)

使用指数法则:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= 2 sin^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= 2 sin^{2} (θ)

= \frac{2 sin ^{2} ( θ )}{cos ( θ )}

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

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} ( θ ) )}{cos ( θ )} + 4 cos (θ)

合并分式

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

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

将项转换为分式:

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

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

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} ( θ ) ) + 4 cos ( θ ) cos ( θ )}{cos ( θ )}

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

- 2 (1 - cos^{2} (θ)) + 4 cos (θ) cos (θ)

4 cos (θ) cos (θ) = 4 cos^{2} (θ)

4 cos (θ) cos (θ)

使用指数法则:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 4 cos^{1 + 1} (θ)

数字相加:

1 + 1 = 2

= 4 cos^{2} (θ)

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

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

乘开

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

- 2 (1 - cos^{2} (θ)) + 4 cos^{2} (θ)

乘开

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

- 2 (1 - cos^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = - 2, b = 1, c = cos^{2} (θ)

= - 2 \cdot 1 - (- 2) cos^{2} (θ)

使用加减运算法则

- (- a) = a

= - 2 \cdot 1 + 2 cos^{2} (θ)

数字相乘:

2 \cdot 1 = 2

= - 2 + 2 cos^{2} (θ)

= - 2 + 2 cos^{2} (θ) + 4 cos^{2} (θ)

同类项相加:

2 cos^{2} (θ) + 4 cos^{2} (θ) = 6 cos^{2} (θ)

= - 2 + 6 cos^{2} (θ)

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

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

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

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

用替代法求解

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

令

: cos (θ) = u

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

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

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

- 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 \cdot 6 ( - 2 )}{2 \cdot 6}

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

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

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 6 \cdot 2 = 48

4 \cdot 6 \cdot 2

数字相乘:

4 \cdot 6 \cdot 2 = 48

= 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 \cdot 6}

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

1 + 7 = 8

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

数字相乘:

2 \cdot 6 = 12

= \frac{8}{12}

约分:

4

= \frac{2}{3}

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

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

使用法则

- (- a) = a

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

数字相减:

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}

二次方程组的解是:

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

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

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

u = cos (θ)

代回

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

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

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

cos (θ) = \frac{2}{3}

使用反三角函数性质

cos (θ) = \frac{2}{3}

cos (θ) = \frac{2}{3}

的通解

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

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

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

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

合并所有解

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

以小数形式表示解

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

作图

流行的例子

3tan(θ)=tan(2θ)

3 tan (θ) = tan (2 θ)

5 cos (x) - 4 = 0

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

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

sin (x) = 14

2sec^2(x)-5sec(x)+2=0,0<= x<= 360

2 sec^{2} (x) - 5 sec (x) + 2 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}