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人気のある三角関数 >

3tan^2(y)=sec(y)+1

解

3 tan^{2} (y) = sec (y) + 1

解

y = 0.72273 \dots + 2 πn, y = 2 π - 0.72273 \dots + 2 πn, y = π + 2 πn

+1

度

y = 41.40962 \dots^{\circ} + 36 0^{\circ} n, y = 318.59037 \dots^{\circ} + 36 0^{\circ} n, y = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

3 tan^{2} (y) = sec (y) + 1

両辺から

sec (y) + 1

を引く

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

三角関数の公式を使用して書き換える

- 1 - sec (y) + 3 tan^{2} (y)

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 (y) + 3 (sec^{2} (y) - 1)

簡素化

- 1 - sec (y) + 3 (sec^{2} (y) - 1) : 3 sec^{2} (y) - sec (y) - 4

- 1 - sec (y) + 3 (sec^{2} (y) - 1)

拡張

3 (sec^{2} (y) - 1) : 3 sec^{2} (y) - 3

3 (sec^{2} (y) - 1)

分配法則を適用する:

a (b - c) = ab - a c

a = 3, b = sec^{2} (y), c = 1

= 3 sec^{2} (y) - 3 \cdot 1

数を乗じる:

3 \cdot 1 = 3

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

= - 1 - sec (y) + 3 sec^{2} (y) - 3

簡素化

- 1 - sec (y) + 3 sec^{2} (y) - 3 : 3 sec^{2} (y) - sec (y) - 4

- 1 - sec (y) + 3 sec^{2} (y) - 3

条件のようなグループ

= - sec (y) + 3 sec^{2} (y) - 1 - 3

数を引く:

- 1 - 3 = - 4

= 3 sec^{2} (y) - sec (y) - 4

= 3 sec^{2} (y) - sec (y) - 4

= 3 sec^{2} (y) - sec (y) - 4

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

置換で解く

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

仮定:

sec (y) = u

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

- 4 - u + 3 u^{2} = 0 : u = \frac{4}{3}, u = - 1

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

標準的な形式で書く

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

3 u^{2} - u - 4 = 0

解くとthe二次式

3 u^{2} - u - 4 = 0

二次Equationの公式:

次の場合:

a = 3, b = - 1, c = - 4

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

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

(- 1)^{2} - 4 \cdot 3 (- 4) = 7

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

規則を適用

- (- a) = a

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

(- 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 3 \cdot 4 = 48

4 \cdot 3 \cdot 4

数を乗じる:

4 \cdot 3 \cdot 4 = 48

= 48

= 1 + 48

数を足す:

1 + 48 = 49

= 49

数を因数に分解する:

49 = 7^{2}

= 7^{2}

累乗根の規則を適用する:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 3}

解を分離する

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 3}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 3}

u = \frac{- ( - 1 ) + 7}{2 \cdot 3} : \frac{4}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{1 + 7}{2 \cdot 3}

数を足す:

1 + 7 = 8

= \frac{8}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{8}{6}

共通因数を約分する:

2

= \frac{4}{3}

u = \frac{- ( - 1 ) - 7}{2 \cdot 3} : - 1

\frac{- ( - 1 ) - 7}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{1 - 7}{2 \cdot 3}

数を引く:

1 - 7 = - 6

= \frac{- 6}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{- 6}{6}

分数の規則を適用する:

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

= - \frac{6}{6}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

u = \frac{4}{3}, u = - 1

代用を戻す

u = sec (y)

sec (y) = \frac{4}{3}, sec (y) = - 1

sec (y) = \frac{4}{3}, sec (y) = - 1

sec (y) = \frac{4}{3} : y = arcsec (\frac{4}{3}) + 2 πn, y = 2 π - arcsec (\frac{4}{3}) + 2 πn

sec (y) = \frac{4}{3}

三角関数の逆数プロパティを適用する

sec (y) = \frac{4}{3}

以下の一般解

sec (y) = \frac{4}{3}

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

y = arcsec (\frac{4}{3}) + 2 πn, y = 2 π - arcsec (\frac{4}{3}) + 2 πn

y = arcsec (\frac{4}{3}) + 2 πn, y = 2 π - arcsec (\frac{4}{3}) + 2 πn

sec (y) = - 1 : y = π + 2 πn

sec (y) = - 1

以下の一般解

sec (y) = - 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}

y = π + 2 πn

y = π + 2 πn

すべての解を組み合わせる

y = arcsec (\frac{4}{3}) + 2 πn, y = 2 π - arcsec (\frac{4}{3}) + 2 πn, y = π + 2 πn

10進法形式で解を証明する

y = 0.72273 \dots + 2 πn, y = 2 π - 0.72273 \dots + 2 πn, y = π + 2 πn

グラフ

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