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人気のある三角関数 >

3tan^2(y)=5sec(y)-1

解

3 tan^{2} (y) = 5 sec (y) - 1

解

y = \frac{π}{3} + 2 πn, y = \frac{5 π}{3} + 2 πn

+1

度

y = 6 0^{\circ} + 36 0^{\circ} n, y = 30 0^{\circ} + 36 0^{\circ} n

解答ステップ

3 tan^{2} (y) = 5 sec (y) - 1

両辺から

5 sec (y) - 1

を引く

3 tan^{2} (y) - 5 sec (y) + 1 = 0

三角関数の公式を使用して書き換える

1 + 3 tan^{2} (y) - 5 sec (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 + 3 (sec^{2} (y) - 1) - 5 sec (y)

簡素化

1 + 3 (sec^{2} (y) - 1) - 5 sec (y) : 3 sec^{2} (y) - 5 sec (y) - 2

1 + 3 (sec^{2} (y) - 1) - 5 sec (y)

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 + 3 sec^{2} (y) - 3 - 5 sec (y)

簡素化

1 + 3 sec^{2} (y) - 3 - 5 sec (y) : 3 sec^{2} (y) - 5 sec (y) - 2

1 + 3 sec^{2} (y) - 3 - 5 sec (y)

条件のようなグループ

= 3 sec^{2} (y) - 5 sec (y) + 1 - 3

数を足す/引く:

1 - 3 = - 2

= 3 sec^{2} (y) - 5 sec (y) - 2

= 3 sec^{2} (y) - 5 sec (y) - 2

= 3 sec^{2} (y) - 5 sec (y) - 2

- 2 + 3 sec^{2} (y) - 5 sec (y) = 0

置換で解く

- 2 + 3 sec^{2} (y) - 5 sec (y) = 0

仮定:

sec (y) = u

- 2 + 3 u^{2} - 5 u = 0

- 2 + 3 u^{2} - 5 u = 0 : u = 2, u = - \frac{1}{3}

- 2 + 3 u^{2} - 5 u = 0

標準的な形式で書く

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

3 u^{2} - 5 u - 2 = 0

解くとthe二次式

3 u^{2} - 5 u - 2 = 0

二次Equationの公式:

次の場合:

a = 3, b = - 5, c = - 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 ( - 2 )}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 (- 2) = 7

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

規則を適用

- (- a) = a

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

指数の規則を適用する:

n

が偶数であれば

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

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 3 \cdot 2

数を乗じる:

4 \cdot 3 \cdot 2 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数を足す:

25 + 24 = 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{- ( - 5 ) \pm 7}{2 \cdot 3}

解を分離する

u_{1} = \frac{- ( - 5 ) + 7}{2 \cdot 3}, u_{2} = \frac{- ( - 5 ) - 7}{2 \cdot 3}

u = \frac{- ( - 5 ) + 7}{2 \cdot 3} : 2

\frac{- ( - 5 ) + 7}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{5 + 7}{2 \cdot 3}

数を足す:

5 + 7 = 12

= \frac{12}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{12}{6}

数を割る:

\frac{12}{6} = 2

= 2

u = \frac{- ( - 5 ) - 7}{2 \cdot 3} : - \frac{1}{3}

\frac{- ( - 5 ) - 7}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{5 - 7}{2 \cdot 3}

数を引く:

5 - 7 = - 2

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

数を乗じる:

2 \cdot 3 = 6

= \frac{- 2}{6}

分数の規則を適用する:

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

= - \frac{2}{6}

共通因数を約分する:

2

= - \frac{1}{3}

二次equationの解:

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

代用を戻す

u = sec (y)

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

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

sec (y) = 2 : y = \frac{π}{3} + 2 πn, y = \frac{5 π}{3} + 2 πn

sec (y) = 2

以下の一般解

sec (y) = 2

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 = \frac{π}{3} + 2 πn, y = \frac{5 π}{3} + 2 πn

y = \frac{π}{3} + 2 πn, y = \frac{5 π}{3} + 2 πn

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

解なし

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

sec (x) \leq - 1 or sec (x) \geq 1

解なし

すべての解を組み合わせる

y = \frac{π}{3} + 2 πn, y = \frac{5 π}{3} + 2 πn

グラフ

人気の例

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