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受欢迎的三角函数 >

16sec^2(θ)-25=0

解答

16 sec^{2} (θ) - 25 = 0

解答

θ = 0.64350 \dots + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn, θ = 2.49809 \dots + 2 πn, θ = - 2.49809 \dots + 2 πn

+1

度数

θ = 36.86989 \dots^{\circ} + 36 0^{\circ} n, θ = 323.13010 \dots^{\circ} + 36 0^{\circ} n, θ = 143.13010 \dots^{\circ} + 36 0^{\circ} n, θ = - 143.13010 \dots^{\circ} + 36 0^{\circ} n

求解步骤

16 sec^{2} (θ) - 25 = 0

用替代法求解

16 sec^{2} (θ) - 25 = 0

令

: sec (θ) = u

16 u^{2} - 25 = 0

16 u^{2} - 25 = 0 : u = \frac{5}{4}, u = - \frac{5}{4}

16 u^{2} - 25 = 0

将

25

到右边

16 u^{2} - 25 = 0

两边加上

25

16 u^{2} - 25 + 25 = 0 + 25

化简

16 u^{2} = 25

16 u^{2} = 25

两边除以

16

16 u^{2} = 25

两边除以

16

\frac{16 u ^{2}}{16} = \frac{25}{16}

化简

u^{2} = \frac{25}{16}

u^{2} = \frac{25}{16}

对于

x^{2} = f (a)

解为

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

u = \frac{25}{16}, u = - \frac{25}{16}

\frac{25}{16} = \frac{5}{4}

\frac{25}{16}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{25}{16}

16 = 4

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

= \frac{25}{4}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{5}{4}

- \frac{25}{16} = - \frac{5}{4}

- \frac{25}{16}

化简

\frac{25}{16} : \frac{5}{4}

\frac{25}{16}

使用根式运算法则:

n \frac{a}{b} = \frac{n a}{n b},

假定

a \geq 0, b \geq 0

= \frac{25}{16}

16 = 4

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

= \frac{25}{4}

25 = 5

25

因式分解数字:

25 = 5^{2}

= 5^{2}

使用根式运算法则:

n a^{n} = a

5^{2} = 5

= 5

= \frac{5}{4}

= - \frac{5}{4}

u = \frac{5}{4}, u = - \frac{5}{4}

u = sec (θ)

代回

sec (θ) = \frac{5}{4}, sec (θ) = - \frac{5}{4}

sec (θ) = \frac{5}{4}, sec (θ) = - \frac{5}{4}

sec (θ) = \frac{5}{4} : θ = arcsec (\frac{5}{4}) + 2 πn, θ = 2 π - arcsec (\frac{5}{4}) + 2 πn

sec (θ) = \frac{5}{4}

使用反三角函数性质

sec (θ) = \frac{5}{4}

sec (θ) = \frac{5}{4}

的通解

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

θ = arcsec (\frac{5}{4}) + 2 πn, θ = 2 π - arcsec (\frac{5}{4}) + 2 πn

θ = arcsec (\frac{5}{4}) + 2 πn, θ = 2 π - arcsec (\frac{5}{4}) + 2 πn

sec (θ) = - \frac{5}{4} : θ = arcsec (- \frac{5}{4}) + 2 πn, θ = - arcsec (- \frac{5}{4}) + 2 πn

sec (θ) = - \frac{5}{4}

使用反三角函数性质

sec (θ) = - \frac{5}{4}

sec (θ) = - \frac{5}{4}

的通解

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

θ = arcsec (- \frac{5}{4}) + 2 πn, θ = - arcsec (- \frac{5}{4}) + 2 πn

θ = arcsec (- \frac{5}{4}) + 2 πn, θ = - arcsec (- \frac{5}{4}) + 2 πn

合并所有解

θ = arcsec (\frac{5}{4}) + 2 πn, θ = 2 π - arcsec (\frac{5}{4}) + 2 πn, θ = arcsec (- \frac{5}{4}) + 2 πn, θ = - arcsec (- \frac{5}{4}) + 2 πn

以小数形式表示解

θ = 0.64350 \dots + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn, θ = 2.49809 \dots + 2 πn, θ = - 2.49809 \dots + 2 πn

作图

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