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受欢迎的三角函数 >

cos(2x)=tan(2x)

解答

cos (2 x) = tan (2 x)

解答

x = \frac{0.66623 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.66623 \dots}{2} + πn

+1

度数

x = 19.08635 \dots^{\circ} + 18 0^{\circ} n, x = 70.91364 \dots^{\circ} + 18 0^{\circ} n

求解步骤

cos (2 x) = tan (2 x)

两边减去

tan (2 x)

cos (2 x) - tan (2 x) = 0

用 sin, cos 表示

cos (2 x) - tan (2 x)

使用基本三角恒等式:

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

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

化简

cos (2 x) - \frac{sin ( 2 x )}{cos ( 2 x )} : \frac{cos ^{2} ( 2 x ) - sin ( 2 x )}{cos ( 2 x )}

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

将项转换为分式:

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

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

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

cos (2 x) cos (2 x) - sin (2 x)

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

cos (2 x) cos (2 x)

使用指数法则:

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

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

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

数字相加:

1 + 1 = 2

= cos^{2} (2 x)

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

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

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

\frac{cos ^{2} ( 2 x ) - sin ( 2 x )}{cos ( 2 x )} = 0

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

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sin (2 x) = u

1 - u - u^{2} = 0

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

1 - u - u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = - 1, b = - 1, c = 1

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

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

(- 1)^{2} - 4 (- 1) \cdot 1 = 5

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

使用法则

- (- a) = a

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

(- 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 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 ( - 1 )}

将解分隔开

u_{1} = \frac{- ( - 1 ) + 5}{2 ( - 1 )}, u_{2} = \frac{- ( - 1 ) - 5}{2 ( - 1 )}

u = \frac{- ( - 1 ) + 5}{2 ( - 1 )} : - \frac{1 + 5}{2}

\frac{- ( - 1 ) + 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 + 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{- 2}

使用分式法则:

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

= - \frac{1 + 5}{2}

u = \frac{- ( - 1 ) - 5}{2 ( - 1 )} : \frac{5 - 1}{2}

\frac{- ( - 1 ) - 5}{2 ( - 1 )}

去除括号:

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

= \frac{1 - 5}{- 2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{- 2}

使用分式法则:

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

1 - 5 = - (5 - 1)

= \frac{5 - 1}{2}

二次方程组的解是:

u = - \frac{1 + 5}{2}, u = \frac{5 - 1}{2}

u = sin (2 x)

代回

sin (2 x) = - \frac{1 + 5}{2}, sin (2 x) = \frac{5 - 1}{2}

sin (2 x) = - \frac{1 + 5}{2}, sin (2 x) = \frac{5 - 1}{2}

sin (2 x) = - \frac{1 + 5}{2} :

无解

sin (2 x) = - \frac{1 + 5}{2}

- 1 \leq sin (x) \leq 1

无解

sin (2 x) = \frac{5 - 1}{2} : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

sin (2 x) = \frac{5 - 1}{2}

使用反三角函数性质

sin (2 x) = \frac{5 - 1}{2}

sin (2 x) = \frac{5 - 1}{2}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

2 x = arcsin (\frac{5 - 1}{2}) + 2 πn, 2 x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

2 x = arcsin (\frac{5 - 1}{2}) + 2 πn, 2 x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

解

2 x = arcsin (\frac{5 - 1}{2}) + 2 πn : x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

2 x = arcsin (\frac{5 - 1}{2}) + 2 πn

两边除以

2

2 x = arcsin (\frac{5 - 1}{2}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

解

2 x = π - arcsin (\frac{5 - 1}{2}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

2 x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{5 - 1}{2}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

合并所有解

x = \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{5 - 1}{2} )}{2} + πn

以小数形式表示解

x = \frac{0.66623 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.66623 \dots}{2} + πn

作图

流行的例子

sin (\frac{θ}{2}) - 1 = 0

2sin(4x)+sqrt(3)=0

2 sin (4 x) + 3 = 0

7sec(x)+7tan(x)=7

7 sec (x) + 7 tan (x) = 7

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

sin (2 x) = - 2 sin (x)

10sec^2(x)+5tan^2(x)-15=0

10 sec^{2} (x) + 5 tan^{2} (x) - 15 = 0