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受欢迎的三角函数 >

2sec^2(x)-tan^4(x)=-1

解答

2 sec^{2} (x) - tan^{4} (x) = - 1

解答

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

+1

度数

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

求解步骤

2 sec^{2} (x) - tan^{4} (x) = - 1

两边减去

- 1

2 sec^{2} (x) - tan^{4} (x) + 1 = 0

使用三角恒等式改写

1 - tan^{4} (x) + 2 sec^{2} (x)

使用毕达哥拉斯恒等式:

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

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

化简

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

1 - tan^{4} (x) + 2 (tan^{2} (x) + 1)

乘开

2 (tan^{2} (x) + 1) : 2 tan^{2} (x) + 2

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

使用分配律:

a (b + c) = ab + a c

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

= 2 tan^{2} (x) + 2 \cdot 1

数字相乘:

2 \cdot 1 = 2

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

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

化简

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

1 - tan^{4} (x) + 2 tan^{2} (x) + 2

对同类项分组

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

数字相加:

1 + 2 = 3

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

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

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

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

用替代法求解

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

令

: tan (x) = u

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

3 - u^{4} + 2 u^{2} = 0 : u = i, u = - i, u = 3, u = - 3

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

改写成标准形式

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

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

用

v = u^{2}

和

v^{2} = u^{4}

改写方程式

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

解

- v^{2} + 2 v + 3 = 0 : v = - 1, v = 3

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 3

数字相乘:

4 \cdot 1 \cdot 3 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

数字相加:

4 + 12 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

v_{1, 2} = \frac{- 2 \pm 4}{2 ( - 1 )}

将解分隔开

v_{1} = \frac{- 2 + 4}{2 ( - 1 )}, v_{2} = \frac{- 2 - 4}{2 ( - 1 )}

v = \frac{- 2 + 4}{2 ( - 1 )} : - 1

\frac{- 2 + 4}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 2 + 4}{- 2 \cdot 1}

数字相加/相减:

- 2 + 4 = 2

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

数字相乘:

2 \cdot 1 = 2

= \frac{2}{- 2}

使用分式法则:

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

= - \frac{2}{2}

使用法则

\frac{a}{a} = 1

= - 1

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

\frac{- 2 - 4}{2 ( - 1 )}

去除括号:

(- a) = - a

= \frac{- 2 - 4}{- 2 \cdot 1}

数字相减:

- 2 - 4 = - 6

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

数字相乘:

2 \cdot 1 = 2

= \frac{- 6}{- 2}

使用分式法则:

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

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

二次方程组的解是:

v = - 1, v = 3

v = - 1, v = 3

代回

v = u^{2}

，求解

u

解

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = - 1, u = - - 1

化简

- 1 : i

- 1

使用虚数运算法则:

- 1 = i

= i

化简

- - 1 : - i

- - 1

使用虚数运算法则:

- 1 = i

= - i

u = i, u = - i

解

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

u = 3, u = - 3

解为

u = i, u = - i, u = 3, u = - 3

u = tan (x)

代回

tan (x) = i, tan (x) = - i, tan (x) = 3, tan (x) = - 3

tan (x) = i, tan (x) = - i, tan (x) = 3, tan (x) = - 3

tan (x) = i :

无解

tan (x) = i

无解

tan (x) = - i :

无解

tan (x) = - i

无解

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

tan (x) = 3

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

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

tan (x) = - 3

tan (x) = - 3

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

合并所有解

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

作图

流行的例子

4 sin (θ) = 2

6 cos^{2} (x) = 3

3sin(x)+sqrt(3)=sin(x)

3 sin (x) + 3 = sin (x)

4sec^2(x)-7tan^2(x)=3

4 sec^{2} (x) - 7 tan^{2} (x) = 3

sin(3x)=cos(2x)

sin (3 x) = cos (2 x)