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人気のある三角関数 >

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

解

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

解

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn, x = π + 2 πn

+1

度

x = 48.18968 \dots^{\circ} + 36 0^{\circ} n, x = 311.81031 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

1

を引く

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

拡張

2 (sec^{2} (x) - 1) : 2 sec^{2} (x) - 2

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

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = sec^{2} (x), c = 1

= 2 sec^{2} (x) - 2 \cdot 1

数を乗じる:

2 \cdot 1 = 2

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

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

簡素化

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

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

条件のようなグループ

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

数を引く:

- 1 - 2 = - 3

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

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

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

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

置換で解く

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

仮定:

sec (x) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

(- 1)^{2} - 4 \cdot 2 (- 3) = 5

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

規則を適用

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

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

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

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 2 \cdot 3 = 24

4 \cdot 2 \cdot 3

数を乗じる:

4 \cdot 2 \cdot 3 = 24

= 24

= 1 + 24

数を足す:

1 + 24 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

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

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

1 + 5 = 6

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

数を乗じる:

2 \cdot 2 = 4

= \frac{6}{4}

共通因数を約分する:

2

= \frac{3}{2}

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

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

規則を適用

- (- a) = a

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

数を引く:

1 - 5 = - 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{4}

分数の規則を適用する:

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

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

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

代用を戻す

u = sec (x)

sec (x) = \frac{3}{2}, sec (x) = - 1

sec (x) = \frac{3}{2}, sec (x) = - 1

sec (x) = \frac{3}{2} : x = arcsec (\frac{3}{2}) + 2 πn, x = 2 π - arcsec (\frac{3}{2}) + 2 πn

sec (x) = \frac{3}{2}

三角関数の逆数プロパティを適用する

sec (x) = \frac{3}{2}

以下の一般解

sec (x) = \frac{3}{2}

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

x = arcsec (\frac{3}{2}) + 2 πn, x = 2 π - arcsec (\frac{3}{2}) + 2 πn

x = arcsec (\frac{3}{2}) + 2 πn, x = 2 π - arcsec (\frac{3}{2}) + 2 πn

sec (x) = - 1 : x = π + 2 πn

sec (x) = - 1

以下の一般解

sec (x) = - 1

sec (x)

2 πn

循環を含む周期性テーブル :

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}

x = π + 2 πn

x = π + 2 πn

すべての解を組み合わせる

x = arcsec (\frac{3}{2}) + 2 πn, x = 2 π - arcsec (\frac{3}{2}) + 2 πn, x = π + 2 πn

10進法形式で解を証明する

x = 0.84106 \dots + 2 πn, x = 2 π - 0.84106 \dots + 2 πn, x = π + 2 πn

グラフ

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