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人気のある三角関数 >

2sec^2(x)-13sec(x)+20=0,2pi<= x<= 4pi

解

2 sec^{2} (x) - 13 sec (x) + 20 = 0, 2 π \leq x \leq 4 π

解

x = 1.31811 \dots + 2 π, x = 4 π - 1.31811 \dots, x = 1.15927 \dots + 2 π, x = 4 π - 1.15927 \dots

+1

度

x = 435.52248 \dots^{\circ}, x = 644.47751 \dots^{\circ}, x = 426.42182 \dots^{\circ}, x = 653.57817 \dots^{\circ}

解答ステップ

2 sec^{2} (x) - 13 sec (x) + 20 = 0, 2 π \leq x \leq 4 π

置換で解く

2 sec^{2} (x) - 13 sec (x) + 20 = 0

仮定:

sec (x) = u

2 u^{2} - 13 u + 20 = 0

2 u^{2} - 13 u + 20 = 0 : u = 4, u = \frac{5}{2}

2 u^{2} - 13 u + 20 = 0

解くとthe二次式

2 u^{2} - 13 u + 20 = 0

二次Equationの公式:

次の場合:

a = 2, b = - 13, c = 20

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 20}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 20}{2 \cdot 2}

(- 13)^{2} - 4 \cdot 2 \cdot 20 = 3

(- 13)^{2} - 4 \cdot 2 \cdot 20

指数の規則を適用する:

n

が偶数であれば

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

(- 13)^{2} = 1 3^{2}

= 1 3^{2} - 4 \cdot 2 \cdot 20

数を乗じる:

4 \cdot 2 \cdot 20 = 160

= 1 3^{2} - 160

1 3^{2} = 169

= 169 - 160

数を引く:

169 - 160 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 13 ) \pm 3}{2 \cdot 2}

解を分離する

u_{1} = \frac{- ( - 13 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 13 ) - 3}{2 \cdot 2}

u = \frac{- ( - 13 ) + 3}{2 \cdot 2} : 4

\frac{- ( - 13 ) + 3}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{13 + 3}{2 \cdot 2}

数を足す:

13 + 3 = 16

= \frac{16}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{16}{4}

数を割る:

\frac{16}{4} = 4

= 4

u = \frac{- ( - 13 ) - 3}{2 \cdot 2} : \frac{5}{2}

\frac{- ( - 13 ) - 3}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{13 - 3}{2 \cdot 2}

数を引く:

13 - 3 = 10

= \frac{10}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{10}{4}

共通因数を約分する:

2

= \frac{5}{2}

二次equationの解:

u = 4, u = \frac{5}{2}

代用を戻す

u = sec (x)

sec (x) = 4, sec (x) = \frac{5}{2}

sec (x) = 4, sec (x) = \frac{5}{2}

sec (x) = 4, 2 π \leq x \leq 4 π : x = arcsec (4) + 2 π, x = 4 π - arcsec (4)

sec (x) = 4, 2 π \leq x \leq 4 π

三角関数の逆数プロパティを適用する

sec (x) = 4

以下の一般解

sec (x) = 4

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

x = arcsec (4) + 2 πn, x = 2 π - arcsec (4) + 2 πn

x = arcsec (4) + 2 πn, x = 2 π - arcsec (4) + 2 πn

範囲の解答

2 π \leq x \leq 4 π

x = arcsec (4) + 2 π, x = 4 π - arcsec (4)

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

sec (x) = \frac{5}{2}, 2 π \leq x \leq 4 π

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

範囲の解答

2 π \leq x \leq 4 π

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 1.31811 \dots + 2 π, x = 4 π - 1.31811 \dots, x = 1.15927 \dots + 2 π, x = 4 π - 1.15927 \dots

グラフ

人気の例

1 = - sin (x)

cos (4 5^{\circ} - x) = \frac{1}{2}

sin(x)-sin^2(x)= 1/4

sin (x) - sin^{2} (x) = \frac{1}{4}

20sin(x)=20cos(x)

20 sin (x) = 20 cos (x)

(-5cot(x))/2+7/3 =-1/6

\frac{- 5 cot ( x )}{2} + \frac{7}{3} = - \frac{1}{6}