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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 = 2 πn, x = 2.30052 \dots + 2 πn, x = - 2.30052 \dots + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 131.81031 \dots^{\circ} + 36 0^{\circ} n, x = - 131.81031 \dots^{\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 = 1, u = - \frac{3}{2}

- 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

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 3)

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 3

数を乗じる:

4 \cdot 2 \cdot 3 = 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} : 1

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

数を足す/引く:

- 1 + 5 = 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

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

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

数を引く:

- 1 - 5 = - 6

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 6}{4}

分数の規則を適用する:

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

= - \frac{6}{4}

共通因数を約分する:

2

= - \frac{3}{2}

二次equationの解:

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

代用を戻す

u = sec (x)

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

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

sec (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 2 πn, x = 2.30052 \dots + 2 πn, x = - 2.30052 \dots + 2 πn

グラフ

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