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受欢迎的三角函数 >

1/(cos(2x))+tan(2x)=3cos(2x)

解答

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

解答

x = \frac{0.72972 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.72972 \dots}{2} + πn

+1

度数

x = 20.90515 \dots^{\circ} + 18 0^{\circ} n, x = 69.09484 \dots^{\circ} + 18 0^{\circ} n

求解步骤

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

两边减去

3 cos (2 x)

\frac{1}{cos ( 2 x )} + tan (2 x) - 3 cos (2 x) = 0

化简

\frac{1}{cos ( 2 x )} + tan (2 x) - 3 cos (2 x) : \frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ^{2} ( 2 x )}{cos ( 2 x )}

\frac{1}{cos ( 2 x )} + tan (2 x) - 3 cos (2 x)

将项转换为分式:

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

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

1 + tan (2 x) cos (2 x) - 3 cos (2 x) cos (2 x) = 1 + tan (2 x) cos (2 x) - 3 cos^{2} (2 x)

1 + tan (2 x) cos (2 x) - 3 cos (2 x) cos (2 x)

3 cos (2 x) cos (2 x) = 3 cos^{2} (2 x)

3 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)

= 3 cos^{1 + 1} (2 x)

数字相加:

1 + 1 = 2

= 3 cos^{2} (2 x)

= 1 + tan (2 x) cos (2 x) - 3 cos^{2} (2 x)

= \frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ^{2} ( 2 x )}{cos ( 2 x )}

\frac{1 + tan ( 2 x ) cos ( 2 x ) - 3 cos ^{2} ( 2 x )}{cos ( 2 x )} = 0

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

1 + tan (2 x) cos (2 x) - 3 cos^{2} (2 x) = 0

用 sin, cos 表示

1 + \frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) - 3 cos^{2} (2 x) = 0

化简

1 + \frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) - 3 cos^{2} (2 x) : 1 + sin (2 x) - 3 cos^{2} (2 x)

1 + \frac{sin ( 2 x )}{cos ( 2 x )} cos (2 x) - 3 cos^{2} (2 x)

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

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

分式相乘:

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

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

约分:

cos (2 x)

= sin (2 x)

= 1 + sin (2 x) - 3 cos^{2} (2 x)

1 + sin (2 x) - 3 cos^{2} (2 x) = 0

两边加上

3 cos^{2} (2 x)

1 + sin (2 x) = 3 cos^{2} (2 x)

两边进行平方

(1 + sin (2 x))^{2} = (3 cos^{2} (2 x))^{2}

两边减去

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

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x) = 0

分解

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x) : (1 + sin (2 x) + 3 cos^{2} (2 x)) (1 + sin (2 x) - 3 cos^{2} (2 x))

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x)

将

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x)

改写为

(1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2}

(1 + sin (2 x))^{2} - 9 cos^{4} (2 x)

将

9

改写为

3^{2}

= (1 + sin (2 x))^{2} - 3^{2} cos^{4} (2 x)

使用指数法则:

a^{b c} = (a^{b})^{c}

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

= (1 + sin (2 x))^{2} - 3^{2} (cos^{2} (2 x))^{2}

使用指数法则:

a^{m} b^{m} = (ab)^{m}

3^{2} (cos^{2} (2 x))^{2} = (3 cos^{2} (2 x))^{2}

= (1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2}

= (1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2}

使用平方差公式:

x^{2} - y^{2} = (x + y) (x - y)

(1 + sin (2 x))^{2} - (3 cos^{2} (2 x))^{2} = ((1 + sin (2 x)) + 3 cos^{2} (2 x)) ((1 + sin (2 x)) - 3 cos^{2} (2 x))

= ((1 + sin (2 x)) + 3 cos^{2} (2 x)) ((1 + sin (2 x)) - 3 cos^{2} (2 x))

整理后得

= (3 cos^{2} (2 x) + sin (2 x) + 1) (sin (2 x) - 3 cos^{2} (2 x) + 1)

(1 + sin (2 x) + 3 cos^{2} (2 x)) (1 + sin (2 x) - 3 cos^{2} (2 x)) = 0

分别求解每个部分

1 + sin (2 x) + 3 cos^{2} (2 x) = 0 or 1 + sin (2 x) - 3 cos^{2} (2 x) = 0

1 + sin (2 x) + 3 cos^{2} (2 x) = 0 : x = \frac{3 π}{4} + πn

1 + sin (2 x) + 3 cos^{2} (2 x) = 0

使用三角恒等式改写

1 + sin (2 x) + 3 cos^{2} (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 x) + 3 (1 - sin^{2} (2 x))

化简

1 + sin (2 x) + 3 (1 - sin^{2} (2 x)) : sin (2 x) - 3 sin^{2} (2 x) + 4

1 + sin (2 x) + 3 (1 - sin^{2} (2 x))

乘开

3 (1 - sin^{2} (2 x)) : 3 - 3 sin^{2} (2 x)

3 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (2 x)

= 3 \cdot 1 - 3 sin^{2} (2 x)

数字相乘:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (2 x)

= 1 + sin (2 x) + 3 - 3 sin^{2} (2 x)

化简

1 + sin (2 x) + 3 - 3 sin^{2} (2 x) : sin (2 x) - 3 sin^{2} (2 x) + 4

1 + sin (2 x) + 3 - 3 sin^{2} (2 x)

对同类项分组

= sin (2 x) - 3 sin^{2} (2 x) + 1 + 3

数字相加:

1 + 3 = 4

= sin (2 x) - 3 sin^{2} (2 x) + 4

= sin (2 x) - 3 sin^{2} (2 x) + 4

= sin (2 x) - 3 sin^{2} (2 x) + 4

4 + sin (2 x) - 3 sin^{2} (2 x) = 0

用替代法求解

4 + sin (2 x) - 3 sin^{2} (2 x) = 0

令

: sin (2 x) = u

4 + u - 3 u^{2} = 0

4 + u - 3 u^{2} = 0 : u = - 1, u = \frac{4}{3}

4 + u - 3 u^{2} = 0

改写成标准形式

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

- 3 u^{2} + u + 4 = 0

使用求根公式求解

- 3 u^{2} + u + 4 = 0

二次方程求根公式:

若

a = - 3, b = 1, c = 4

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 4}{2 ( - 3 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 4}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 4 = 7

1^{2} - 4 (- 3) \cdot 4

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 3) \cdot 4

使用法则

- (- a) = a

= 1 + 4 \cdot 3 \cdot 4

数字相乘:

4 \cdot 3 \cdot 4 = 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 ( - 3 )}

将解分隔开

u_{1} = \frac{- 1 + 7}{2 ( - 3 )}, u_{2} = \frac{- 1 - 7}{2 ( - 3 )}

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

\frac{- 1 + 7}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 1 + 7}{- 2 \cdot 3}

数字相加/相减:

- 1 + 7 = 6

= \frac{6}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{6}{- 6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 1 - 7}{2 ( - 3 )} : \frac{4}{3}

\frac{- 1 - 7}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 1 - 7}{- 2 \cdot 3}

数字相减:

- 1 - 7 = - 8

= \frac{- 8}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 8}{- 6}

使用分式法则:

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

= \frac{8}{6}

约分:

2

= \frac{4}{3}

二次方程组的解是:

u = - 1, u = \frac{4}{3}

u = sin (2 x)

代回

sin (2 x) = - 1, sin (2 x) = \frac{4}{3}

sin (2 x) = - 1, sin (2 x) = \frac{4}{3}

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

sin (2 x) = - 1

sin (2 x) = - 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

解

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

sin (2 x) = \frac{4}{3} :

无解

sin (2 x) = \frac{4}{3}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{3 π}{4} + πn

1 + sin (2 x) - 3 cos^{2} (2 x) = 0 : x = \frac{arcsin ( \frac{2}{3} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + πn, x = \frac{3 π}{4} + πn

1 + sin (2 x) - 3 cos^{2} (2 x) = 0

使用三角恒等式改写

1 + sin (2 x) - 3 cos^{2} (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 x) - 3 (1 - sin^{2} (2 x))

化简

1 + sin (2 x) - 3 (1 - sin^{2} (2 x)) : 3 sin^{2} (2 x) + sin (2 x) - 2

1 + sin (2 x) - 3 (1 - sin^{2} (2 x))

乘开

- 3 (1 - sin^{2} (2 x)) : - 3 + 3 sin^{2} (2 x)

- 3 (1 - sin^{2} (2 x))

使用分配律:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (2 x)

= - 3 \cdot 1 - (- 3) sin^{2} (2 x)

使用加减运算法则

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (2 x)

数字相乘:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (2 x)

= 1 + sin (2 x) - 3 + 3 sin^{2} (2 x)

化简

1 + sin (2 x) - 3 + 3 sin^{2} (2 x) : 3 sin^{2} (2 x) + sin (2 x) - 2

1 + sin (2 x) - 3 + 3 sin^{2} (2 x)

对同类项分组

= sin (2 x) + 3 sin^{2} (2 x) + 1 - 3

数字相加/相减:

1 - 3 = - 2

= 3 sin^{2} (2 x) + sin (2 x) - 2

= 3 sin^{2} (2 x) + sin (2 x) - 2

= 3 sin^{2} (2 x) + sin (2 x) - 2

- 2 + sin (2 x) + 3 sin^{2} (2 x) = 0

用替代法求解

- 2 + sin (2 x) + 3 sin^{2} (2 x) = 0

令

: sin (2 x) = u

- 2 + u + 3 u^{2} = 0

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

- 2 + u + 3 u^{2} = 0

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 3, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

1^{2} - 4 \cdot 3 (- 2) = 5

1^{2} - 4 \cdot 3 (- 2)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 3 (- 2)

使用法则

- (- a) = a

= 1 + 4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 1 + 24

数字相加:

1 + 24 = 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{- 1 \pm 5}{2 \cdot 3}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 3}, u_{2} = \frac{- 1 - 5}{2 \cdot 3}

u = \frac{- 1 + 5}{2 \cdot 3} : \frac{2}{3}

\frac{- 1 + 5}{2 \cdot 3}

数字相加/相减:

- 1 + 5 = 4

= \frac{4}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{4}{6}

约分:

2

= \frac{2}{3}

u = \frac{- 1 - 5}{2 \cdot 3} : - 1

\frac{- 1 - 5}{2 \cdot 3}

数字相减:

- 1 - 5 = - 6

= \frac{- 6}{2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 6}{6}

使用分式法则:

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

= - \frac{6}{6}

使用法则

\frac{a}{a} = 1

= - 1

二次方程组的解是:

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

u = sin (2 x)

代回

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

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

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

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

使用反三角函数性质

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

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

的通解

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

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

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

解

2 x = arcsin (\frac{2}{3}) + 2 πn : x = \frac{arcsin ( \frac{2}{3} )}{2} + πn

2 x = arcsin (\frac{2}{3}) + 2 πn

两边除以

2

2 x = arcsin (\frac{2}{3}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{2}{3} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{2}{3} )}{2} + πn

x = \frac{arcsin ( \frac{2}{3} )}{2} + πn

解

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

2 x = π - arcsin (\frac{2}{3}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{2}{3}) + 2 πn

两边除以

2

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

化简

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

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

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

sin (2 x) = - 1 : x = \frac{3 π}{4} + πn

sin (2 x) = - 1

sin (2 x) = - 1

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

解

2 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

2 x = \frac{3 π}{2} + 2 πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2} : \frac{3 π}{4} + πn

\frac{\frac{3 π}{2}}{2} + \frac{2 πn}{2}

\frac{\frac{3 π}{2}}{2} = \frac{3 π}{4}

\frac{\frac{3 π}{2}}{2}

使用分式法则:

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

= \frac{3 π}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

合并所有解

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

合并所有解

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

将解代入原方程进行验证

将它们代入

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

检验解是否符合

去除与方程不符的解。

检验

\frac{3 π}{4} + πn

的解:

假

\frac{3 π}{4} + πn

代入

n = 1

\frac{3 π}{4} + π 1

对于

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

代入

x = \frac{3 π}{4} + π 1

\frac{1}{cos ( 2 ( \frac{3 π}{4} + π 1 ) )} + tan (2 (\frac{3 π}{4} + π 1)) = 3 cos (2 (\frac{3 π}{4} + π 1))

未定义

\Rightarrow 假

检验

\frac{arcsin ( \frac{2}{3} )}{2} + πn

的解:

真

\frac{arcsin ( \frac{2}{3} )}{2} + πn

代入

n = 1

\frac{arcsin ( \frac{2}{3} )}{2} + π 1

对于

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

代入

x = \frac{arcsin ( \frac{2}{3} )}{2} + π 1

\frac{1}{cos ( 2 ( \frac{a r c s i n ( \frac{2}{3} )}{2} + π 1 ) )} + tan (2 (\frac{arcsin ( \frac{2}{3} )}{2} + π 1)) = 3 cos (2 (\frac{arcsin ( \frac{2}{3} )}{2} + π 1))

整理后得

2.23606 \dots = 2.23606 \dots

\Rightarrow 真

检验

\frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + πn

的解:

真

\frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + πn

代入

n = 1

\frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + π 1

对于

\frac{1}{cos ( 2 x )} + tan (2 x) = 3 cos (2 x)

代入

x = \frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + π 1

\frac{1}{cos ( 2 ( \frac{π}{2} - \frac{a r c s i n ( \frac{2}{3} )}{2} + π 1 ) )} + tan (2 (\frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + π 1)) = 3 cos (2 (\frac{π}{2} - \frac{arcsin ( \frac{2}{3} )}{2} + π 1))

整理后得

- 2.23606 \dots = - 2.23606 \dots

\Rightarrow 真

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

以小数形式表示解

x = \frac{0.72972 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.72972 \dots}{2} + πn

作图

流行的例子

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

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

sin (2 t) = sin (t)

cos(x)=(17.6)/(26)

cos (x) = \frac{17.6}{26}

1+4cos(θ)=sqrt(3)sin(θ),0<= θ<= 2pi

1 + 4 cos (θ) = 3 sin (θ), 0 \leq θ \leq 2 π

tan^2(x)+cot^2(x)=1

tan^{2} (x) + cot^{2} (x) = 1