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受欢迎的三角函数 >

tan^2(x)=sec(x)

解答

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

解答

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

+1

度数

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

sec (x)

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

使用三角恒等式改写

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

使用毕达哥拉斯恒等式:

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

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

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

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

用替代法求解

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

令

: sec (x) = u

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

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

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

改写成标准形式

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

u^{2} - u - 1 = 0

使用求根公式求解

u^{2} - u - 1 = 0

二次方程求根公式:

若

a = 1, b = - 1, c = - 1

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

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

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

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

使用法则

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

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

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

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

将解分隔开

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

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

\frac{- ( - 1 ) + 5}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

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

\frac{- ( - 1 ) - 5}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次方程组的解是:

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

u = sec (x)

代回

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

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

sec (x) = \frac{1 + 5}{2} : x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

sec (x) = \frac{1 + 5}{2}

使用反三角函数性质

sec (x) = \frac{1 + 5}{2}

sec (x) = \frac{1 + 5}{2}

的通解

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

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

sec (x) = \frac{1 - 5}{2} :

无解

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

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

无解

合并所有解

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

以小数形式表示解

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

作图

流行的例子

solvefor x,sin(x)-sin(2x)=0

so l v e f or x, sin (x) - sin (2 x) = 0

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sin(6x)=cos(3x)

sin (6 x) = cos (3 x)

sin (x) = \frac{- 3}{5}

2cos(6x)+sqrt(3)=0

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