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受欢迎的 >

1/(sec^2(x))+1/4 = 2/(csc(x))

解答

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

解答

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度数

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

求解步骤

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

两边减去

\frac{2}{csc ( x )}

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} - \frac{2}{csc ( x )} = 0

化简

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} - \frac{2}{csc ( x )} : \frac{4 csc ( x ) + sec ^{2} ( x ) csc ( x ) - 8 sec ^{2} ( x )}{4 sec ^{2} ( x ) csc ( x )}

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

sec^{2} (x), 4, csc (x)

的最小公倍数:

4 sec^{2} (x) csc (x)

sec^{2} (x), 4, csc (x)

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

= 4 sec^{2} (x) csc (x)

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

4 sec^{2} (x) csc (x)

对于

\frac{1}{sec ^{2} ( x )} :

将分母和分子乘以

4 csc (x)

\frac{1}{sec ^{2} ( x )} = \frac{1 \cdot 4 csc ( x )}{sec ^{2} ( x ) \cdot 4 csc ( x )} = \frac{4 csc ( x )}{4 sec ^{2} ( x ) csc ( x )}

对于

\frac{1}{4} :

将分母和分子乘以

sec^{2} (x) csc (x)

\frac{1}{4} = \frac{1 \cdot sec ^{2} ( x ) csc ( x )}{4 sec ^{2} ( x ) csc ( x )} = \frac{sec ^{2} ( x ) csc ( x )}{4 sec ^{2} ( x ) csc ( x )}

对于

\frac{2}{csc ( x )} :

将分母和分子乘以

4 sec^{2} (x)

\frac{2}{csc ( x )} = \frac{2 \cdot 4 sec ^{2} ( x )}{csc ( x ) \cdot 4 sec ^{2} ( x )} = \frac{8 sec ^{2} ( x )}{4 sec ^{2} ( x ) csc ( x )}

= \frac{4 csc ( x )}{4 sec ^{2} ( x ) csc ( x )} + \frac{sec ^{2} ( x ) csc ( x )}{4 sec ^{2} ( x ) csc ( x )} - \frac{8 sec ^{2} ( x )}{4 sec ^{2} ( x ) csc ( 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{4 csc ( x ) + sec ^{2} ( x ) csc ( x ) - 8 sec ^{2} ( x )}{4 sec ^{2} ( x ) csc ( x )}

\frac{4 csc ( x ) + sec ^{2} ( x ) csc ( x ) - 8 sec ^{2} ( x )}{4 sec ^{2} ( x ) csc ( x )} = 0

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

4 csc (x) + sec^{2} (x) csc (x) - 8 sec^{2} (x) = 0

用 sin, cos 表示

4 \cdot \frac{1}{sin ( x )} + (\frac{1}{cos ( x )})^{2} \frac{1}{sin ( x )} - 8 (\frac{1}{cos ( x )})^{2} = 0

化简

4 \cdot \frac{1}{sin ( x )} + (\frac{1}{cos ( x )})^{2} \frac{1}{sin ( x )} - 8 (\frac{1}{cos ( x )})^{2} : \frac{4 cos ^{2} ( x ) + 1 - 8 sin ( x )}{cos ^{2} ( x ) sin ( x )}

4 \cdot \frac{1}{sin ( x )} + (\frac{1}{cos ( x )})^{2} \frac{1}{sin ( x )} - 8 (\frac{1}{cos ( x )})^{2}

4 \cdot \frac{1}{sin ( x )} = \frac{4}{sin ( x )}

4 \cdot \frac{1}{sin ( x )}

分式相乘:

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

= \frac{1 \cdot 4}{sin ( x )}

数字相乘:

1 \cdot 4 = 4

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

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

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

分式相乘:

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

= \frac{1 \cdot ( \frac{1}{c o s ( x )} ) ^{2}}{sin ( x )}

乘以:

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

= \frac{( \frac{1}{c o s ( x )} ) ^{2}}{sin ( x )}

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

(\frac{1}{cos ( x )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= \frac{\frac{1}{c o s ^{2} ( x )}}{sin ( x )}

使用分式法则:

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

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

8 (\frac{1}{cos ( x )})^{2} = \frac{8}{cos ^{2} ( x )}

8 (\frac{1}{cos ( x )})^{2}

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

(\frac{1}{cos ( x )})^{2}

使用指数法则:

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

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

使用法则

1^{a} = 1

1^{2} = 1

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

= 8 \cdot \frac{1}{cos ^{2} ( x )}

分式相乘:

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

= \frac{1 \cdot 8}{cos ^{2} ( x )}

数字相乘:

1 \cdot 8 = 8

= \frac{8}{cos ^{2} ( x )}

= \frac{4}{sin ( x )} + \frac{1}{cos ^{2} ( x ) sin ( x )} - \frac{8}{cos ^{2} ( x )}

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

的最小公倍数:

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

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

最小公倍数 (LCM)

计算出由至少在以下一个因式表达式中出现的因子组成的表达式

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

根据最小公倍数调整分式

将每个分子乘以其分母转变为最小公倍数所要乘以的同一数值

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

对于

\frac{4}{sin ( x )} :

将分母和分子乘以

cos^{2} (x)

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

对于

\frac{8}{cos ^{2} ( x )} :

将分母和分子乘以

sin (x)

\frac{8}{cos ^{2} ( x )} = \frac{8 sin ( x )}{cos ^{2} ( x ) sin ( x )}

= \frac{4 cos ^{2} ( x )}{sin ( x ) cos ^{2} ( x )} + \frac{1}{cos ^{2} ( x ) sin ( x )} - \frac{8 sin ( x )}{cos ^{2} ( x ) sin ( 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{4 cos ^{2} ( x ) + 1 - 8 sin ( x )}{cos ^{2} ( x ) sin ( x )}

\frac{4 cos ^{2} ( x ) + 1 - 8 sin ( x )}{cos ^{2} ( x ) sin ( x )} = 0

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

4 cos^{2} (x) + 1 - 8 sin (x) = 0

两边加上

8 sin (x)

4 cos^{2} (x) + 1 = 8 sin (x)

两边进行平方

(4 cos^{2} (x) + 1)^{2} = (8 sin (x))^{2}

两边减去

(8 sin (x))^{2}

(4 cos^{2} (x) + 1)^{2} - 64 sin^{2} (x) = 0

分解

(4 cos^{2} (x) + 1)^{2} - 64 sin^{2} (x) : (4 cos^{2} (x) + 1 + 8 sin (x)) (4 cos^{2} (x) + 1 - 8 sin (x))

(4 cos^{2} (x) + 1)^{2} - 64 sin^{2} (x)

将

(4 cos^{2} (x) + 1)^{2} - 64 sin^{2} (x)

改写为

(4 cos^{2} (x) + 1)^{2} - (8 sin (x))^{2}

(4 cos^{2} (x) + 1)^{2} - 64 sin^{2} (x)

将

64

改写为

8^{2}

= (4 cos^{2} (x) + 1)^{2} - 8^{2} sin^{2} (x)

使用指数法则:

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

8^{2} sin^{2} (x) = (8 sin (x))^{2}

= (4 cos^{2} (x) + 1)^{2} - (8 sin (x))^{2}

= (4 cos^{2} (x) + 1)^{2} - (8 sin (x))^{2}

使用平方差公式:

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

(4 cos^{2} (x) + 1)^{2} - (8 sin (x))^{2} = ((4 cos^{2} (x) + 1) + 8 sin (x)) ((4 cos^{2} (x) + 1) - 8 sin (x))

= ((4 cos^{2} (x) + 1) + 8 sin (x)) ((4 cos^{2} (x) + 1) - 8 sin (x))

整理后得

= (4 cos^{2} (x) + 8 sin (x) + 1) (4 cos^{2} (x) - 8 sin (x) + 1)

(4 cos^{2} (x) + 1 + 8 sin (x)) (4 cos^{2} (x) + 1 - 8 sin (x)) = 0

分别求解每个部分

4 cos^{2} (x) + 1 + 8 sin (x) = 0 or 4 cos^{2} (x) + 1 - 8 sin (x) = 0

4 cos^{2} (x) + 1 + 8 sin (x) = 0 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

4 cos^{2} (x) + 1 + 8 sin (x) = 0

使用三角恒等式改写

1 + 4 cos^{2} (x) + 8 sin (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 + 4 (1 - sin^{2} (x)) + 8 sin (x)

化简

1 + 4 (1 - sin^{2} (x)) + 8 sin (x) : 8 sin (x) - 4 sin^{2} (x) + 5

1 + 4 (1 - sin^{2} (x)) + 8 sin (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

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

= 1 + 4 - 4 sin^{2} (x) + 8 sin (x)

数字相加:

1 + 4 = 5

= 8 sin (x) - 4 sin^{2} (x) + 5

= 8 sin (x) - 4 sin^{2} (x) + 5

5 - 4 sin^{2} (x) + 8 sin (x) = 0

用替代法求解

5 - 4 sin^{2} (x) + 8 sin (x) = 0

令

: sin (x) = u

5 - 4 u^{2} + 8 u = 0

5 - 4 u^{2} + 8 u = 0 : u = - \frac{1}{2}, u = \frac{5}{2}

5 - 4 u^{2} + 8 u = 0

改写成标准形式

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

- 4 u^{2} + 8 u + 5 = 0

使用求根公式求解

- 4 u^{2} + 8 u + 5 = 0

二次方程求根公式:

若

a = - 4, b = 8, c = 5

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

8^{2} - 4 (- 4) \cdot 5 = 12

8^{2} - 4 (- 4) \cdot 5

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 4 \cdot 5

数字相乘:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

数字相加:

64 + 80 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- 8 \pm 12}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- 8 + 12}{2 ( - 4 )}, u_{2} = \frac{- 8 - 12}{2 ( - 4 )}

u = \frac{- 8 + 12}{2 ( - 4 )} : - \frac{1}{2}

\frac{- 8 + 12}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 8 + 12}{- 2 \cdot 4}

数字相加/相减:

- 8 + 12 = 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{4}{- 8}

使用分式法则:

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

= - \frac{4}{8}

约分:

4

= - \frac{1}{2}

u = \frac{- 8 - 12}{2 ( - 4 )} : \frac{5}{2}

\frac{- 8 - 12}{2 ( - 4 )}

去除括号:

(- a) = - a

= \frac{- 8 - 12}{- 2 \cdot 4}

数字相减:

- 8 - 12 = - 20

= \frac{- 20}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{- 20}{- 8}

使用分式法则:

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

= \frac{20}{8}

约分:

4

= \frac{5}{2}

二次方程组的解是:

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

u = sin (x)

代回

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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

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}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

无解

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

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

4 cos^{2} (x) + 1 - 8 sin (x) = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

4 cos^{2} (x) + 1 - 8 sin (x) = 0

使用三角恒等式改写

1 + 4 cos^{2} (x) - 8 sin (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 + 4 (1 - sin^{2} (x)) - 8 sin (x)

化简

1 + 4 (1 - sin^{2} (x)) - 8 sin (x) : - 4 sin^{2} (x) - 8 sin (x) + 5

1 + 4 (1 - sin^{2} (x)) - 8 sin (x)

乘开

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

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

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数字相乘:

4 \cdot 1 = 4

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

= 1 + 4 - 4 sin^{2} (x) - 8 sin (x)

数字相加:

1 + 4 = 5

= - 4 sin^{2} (x) - 8 sin (x) + 5

= - 4 sin^{2} (x) - 8 sin (x) + 5

5 - 4 sin^{2} (x) - 8 sin (x) = 0

用替代法求解

5 - 4 sin^{2} (x) - 8 sin (x) = 0

令

: sin (x) = u

5 - 4 u^{2} - 8 u = 0

5 - 4 u^{2} - 8 u = 0 : u = - \frac{5}{2}, u = \frac{1}{2}

5 - 4 u^{2} - 8 u = 0

改写成标准形式

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

- 4 u^{2} - 8 u + 5 = 0

使用求根公式求解

- 4 u^{2} - 8 u + 5 = 0

二次方程求根公式:

若

a = - 4, b = - 8, c = 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 8)^{2} - 4 (- 4) \cdot 5 = 12

(- 8)^{2} - 4 (- 4) \cdot 5

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 4 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 4 \cdot 5

数字相乘:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

数字相加:

64 + 80 = 144

= 144

因式分解数字:

144 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 ( - 4 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 12}{2 ( - 4 )}, u_{2} = \frac{- ( - 8 ) - 12}{2 ( - 4 )}

u = \frac{- ( - 8 ) + 12}{2 ( - 4 )} : - \frac{5}{2}

\frac{- ( - 8 ) + 12}{2 ( - 4 )}

去除括号:

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

= \frac{8 + 12}{- 2 \cdot 4}

数字相加:

8 + 12 = 20

= \frac{20}{- 2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{20}{- 8}

使用分式法则:

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

= - \frac{20}{8}

约分:

4

= - \frac{5}{2}

u = \frac{- ( - 8 ) - 12}{2 ( - 4 )} : \frac{1}{2}

\frac{- ( - 8 ) - 12}{2 ( - 4 )}

去除括号:

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

= \frac{8 - 12}{- 2 \cdot 4}

数字相减:

8 - 12 = - 4

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

数字相乘:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

使用分式法则:

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

= \frac{4}{8}

约分:

4

= \frac{1}{2}

二次方程组的解是:

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

u = sin (x)

代回

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

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

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

无解

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

- 1 \leq sin (x) \leq 1

无解

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

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

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}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

合并所有解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

将解代入原方程进行验证

将它们代入

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

检验解是否符合

去除与方程不符的解。

检验

\frac{7 π}{6} + 2 πn

的解:

假

\frac{7 π}{6} + 2 πn

代入

n = 1

\frac{7 π}{6} + 2 π 1

对于

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

代入

x = \frac{7 π}{6} + 2 π 1

\frac{1}{sec ^{2} ( \frac{7 π}{6} + 2 π 1 )} + \frac{1}{4} = \frac{2}{csc ( \frac{7 π}{6} + 2 π 1 )}

整理后得

1 = - 1

\Rightarrow 假

检验

\frac{11 π}{6} + 2 πn

的解:

假

\frac{11 π}{6} + 2 πn

代入

n = 1

\frac{11 π}{6} + 2 π 1

对于

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

代入

x = \frac{11 π}{6} + 2 π 1

\frac{1}{sec ^{2} ( \frac{11 π}{6} + 2 π 1 )} + \frac{1}{4} = \frac{2}{csc ( \frac{11 π}{6} + 2 π 1 )}

整理后得

1 = - 1

\Rightarrow 假

检验

\frac{π}{6} + 2 πn

的解:

真

\frac{π}{6} + 2 πn

代入

n = 1

\frac{π}{6} + 2 π 1

对于

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

代入

x = \frac{π}{6} + 2 π 1

\frac{1}{sec ^{2} ( \frac{π}{6} + 2 π 1 )} + \frac{1}{4} = \frac{2}{csc ( \frac{π}{6} + 2 π 1 )}

整理后得

1 = 1

\Rightarrow 真

检验

\frac{5 π}{6} + 2 πn

的解:

真

\frac{5 π}{6} + 2 πn

代入

n = 1

\frac{5 π}{6} + 2 π 1

对于

\frac{1}{sec ^{2} ( x )} + \frac{1}{4} = \frac{2}{csc ( x )}

代入

x = \frac{5 π}{6} + 2 π 1

\frac{1}{sec ^{2} ( \frac{5 π}{6} + 2 π 1 )} + \frac{1}{4} = \frac{2}{csc ( \frac{5 π}{6} + 2 π 1 )}

整理后得

1 = 1

\Rightarrow 真

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

流行的例子

cos(x/3)=(sqrt(3))/2

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

tan (θ) = 1.535

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

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

1+cos(θ)=sqrt(3)-sin(θ)

1 + cos (θ) = 3 - sin (θ)

3 tan^{2} (θ) - 9 = 0