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受欢迎的三角函数 >

tan^2(θ)-2sec(θ)=2

解答

tan^{2} (θ) - 2 sec (θ) = 2

解答

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = π + 2 πn

+1

度数

θ = 70.52877 \dots^{\circ} + 36 0^{\circ} n, θ = 289.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

tan^{2} (θ) - 2 sec (θ) = 2

两边减去

2

tan^{2} (θ) - 2 sec (θ) - 2 = 0

使用三角恒等式改写

- 2 + tan^{2} (θ) - 2 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

= - 2 + sec^{2} (θ) - 1 - 2 sec (θ)

化简

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

- 2 + sec^{2} (θ) - 1 - 2 sec (θ)

对同类项分组

= sec^{2} (θ) - 2 sec (θ) - 2 - 1

数字相减:

- 2 - 1 = - 3

= sec^{2} (θ) - 2 sec (θ) - 3

= sec^{2} (θ) - 2 sec (θ) - 3

- 3 + sec^{2} (θ) - 2 sec (θ) = 0

用替代法求解

- 3 + sec^{2} (θ) - 2 sec (θ) = 0

令

: sec (θ) = u

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

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

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

改写成标准形式

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

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

使用求根公式求解

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

二次方程求根公式:

若

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

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

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

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

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 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

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

将解分隔开

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

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

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

使用法则

- (- a) = a

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

数字相加:

2 + 4 = 6

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

数字相乘:

2 \cdot 1 = 2

= \frac{6}{2}

数字相除:

\frac{6}{2} = 3

= 3

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

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

使用法则

- (- a) = a

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

数字相减:

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

二次方程组的解是:

u = 3, u = - 1

u = sec (θ)

代回

sec (θ) = 3, sec (θ) = - 1

sec (θ) = 3, sec (θ) = - 1

sec (θ) = 3 : θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn

sec (θ) = 3

使用反三角函数性质

sec (θ) = 3

sec (θ) = 3

的通解

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn

θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn

sec (θ) = - 1 : θ = π + 2 πn

sec (θ) = - 1

sec (θ) = - 1

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 = π + 2 πn

θ = π + 2 πn

合并所有解

θ = arcsec (3) + 2 πn, θ = 2 π - arcsec (3) + 2 πn, θ = π + 2 πn

以小数形式表示解

θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn, θ = π + 2 πn

作图

流行的例子

-sin(B)-4=3sin(B)-5

- sin (B) - 4 = 3 sin (B) - 5

cos^2(x)-7cos(x)-1=0

cos^{2} (x) - 7 cos (x) - 1 = 0

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

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

0 = sin (θ)

solvefor t,x=sin(t)

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