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受欢迎的三角函数 >

5sec(θ)-2sec^2(θ)=tan^2(θ)-1

解答

5 sec (θ) - 2 sec^{2} (θ) = tan^{2} (θ) - 1

解答

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

+1

度数

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n

求解步骤

5 sec (θ) - 2 sec^{2} (θ) = tan^{2} (θ) - 1

两边减去

tan^{2} (θ) - 1

5 sec (θ) - 2 sec^{2} (θ) - tan^{2} (θ) + 1 = 0

使用三角恒等式改写

1 - tan^{2} (θ) - 2 sec^{2} (θ) + 5 sec (θ)

使用毕达哥拉斯恒等式:

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

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

= 1 - (sec^{2} (θ) - 1) - 2 sec^{2} (θ) + 5 sec (θ)

化简

1 - (sec^{2} (θ) - 1) - 2 sec^{2} (θ) + 5 sec (θ) : 5 sec (θ) - 3 sec^{2} (θ) + 2

1 - (sec^{2} (θ) - 1) - 2 sec^{2} (θ) + 5 sec (θ)

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

- (sec^{2} (θ) - 1)

打开括号

= - (sec^{2} (θ)) - (- 1)

使用加减运算法则

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

= - sec^{2} (θ) + 1

= 1 - sec^{2} (θ) + 1 - 2 sec^{2} (θ) + 5 sec (θ)

化简

1 - sec^{2} (θ) + 1 - 2 sec^{2} (θ) + 5 sec (θ) : 5 sec (θ) - 3 sec^{2} (θ) + 2

1 - sec^{2} (θ) + 1 - 2 sec^{2} (θ) + 5 sec (θ)

对同类项分组

= - sec^{2} (θ) - 2 sec^{2} (θ) + 5 sec (θ) + 1 + 1

同类项相加:

- sec^{2} (θ) - 2 sec^{2} (θ) = - 3 sec^{2} (θ)

= - 3 sec^{2} (θ) + 5 sec (θ) + 1 + 1

数字相加:

1 + 1 = 2

= 5 sec (θ) - 3 sec^{2} (θ) + 2

= 5 sec (θ) - 3 sec^{2} (θ) + 2

= 5 sec (θ) - 3 sec^{2} (θ) + 2

2 - 3 sec^{2} (θ) + 5 sec (θ) = 0

用替代法求解

2 - 3 sec^{2} (θ) + 5 sec (θ) = 0

令

: sec (θ) = u

2 - 3 u^{2} + 5 u = 0

2 - 3 u^{2} + 5 u = 0 : u = - \frac{1}{3}, u = 2

2 - 3 u^{2} + 5 u = 0

改写成标准形式

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

- 3 u^{2} + 5 u + 2 = 0

使用求根公式求解

- 3 u^{2} + 5 u + 2 = 0

二次方程求根公式:

若

a = - 3, b = 5, c = 2

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) \cdot 2}{2 ( - 3 )}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 3 ) \cdot 2}{2 ( - 3 )}

5^{2} - 4 (- 3) \cdot 2 = 7

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

使用法则

- (- a) = a

= 5^{2} + 4 \cdot 3 \cdot 2

数字相乘:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数字相加:

25 + 24 = 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{- 5 \pm 7}{2 ( - 3 )}

将解分隔开

u_{1} = \frac{- 5 + 7}{2 ( - 3 )}, u_{2} = \frac{- 5 - 7}{2 ( - 3 )}

u = \frac{- 5 + 7}{2 ( - 3 )} : - \frac{1}{3}

\frac{- 5 + 7}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 5 + 7}{- 2 \cdot 3}

数字相加/相减:

- 5 + 7 = 2

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

数字相乘:

2 \cdot 3 = 6

= \frac{2}{- 6}

使用分式法则:

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

= - \frac{2}{6}

约分:

2

= - \frac{1}{3}

u = \frac{- 5 - 7}{2 ( - 3 )} : 2

\frac{- 5 - 7}{2 ( - 3 )}

去除括号:

(- a) = - a

= \frac{- 5 - 7}{- 2 \cdot 3}

数字相减:

- 5 - 7 = - 12

= \frac{- 12}{- 2 \cdot 3}

数字相乘:

2 \cdot 3 = 6

= \frac{- 12}{- 6}

使用分式法则:

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

= \frac{12}{6}

数字相除:

\frac{12}{6} = 2

= 2

二次方程组的解是:

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

u = sec (θ)

代回

sec (θ) = - \frac{1}{3}, sec (θ) = 2

sec (θ) = - \frac{1}{3}, sec (θ) = 2

sec (θ) = - \frac{1}{3} :

无解

sec (θ) = - \frac{1}{3}

sec (x) \leq - 1 or sec (x) \geq 1

无解

sec (θ) = 2 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sec (θ) = 2

sec (θ) = 2

的通解

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

合并所有解

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

作图

流行的例子

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