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人気のある三角関数 >

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

解

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

解

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

+1

度

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

1

を引く

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

簡素化

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

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

条件のようなグループ

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

数を引く:

- 1 - 1 = - 2

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

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

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

置換で解く

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

仮定:

sec (x) = u

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

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

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

標準的な形式で書く

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

u^{2} - u - 2 = 0

解くとthe二次式

u^{2} - u - 2 = 0

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

- (- a) = a

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

(- 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 1 \cdot 2 = 8

4 \cdot 1 \cdot 2

数を乗じる:

4 \cdot 1 \cdot 2 = 8

= 8

= 1 + 8

数を足す:

1 + 8 = 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{- ( - 1 ) \pm 3}{2 \cdot 1}

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

1 + 3 = 4

= \frac{4}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{4}{2}

数を割る:

\frac{4}{2} = 2

= 2

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

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

規則を適用

- (- a) = a

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

数を引く:

1 - 3 = - 2

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

数を乗じる:

2 \cdot 1 = 2

= \frac{- 2}{2}

分数の規則を適用する:

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

= - \frac{2}{2}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

u = 2, u = - 1

代用を戻す

u = sec (x)

sec (x) = 2, sec (x) = - 1

sec (x) = 2, sec (x) = - 1

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

sec (x) = 2

以下の一般解

sec (x) = 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}

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

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

グラフ

人気の例

sin(θ)=(sqrt(3))/2

sin (θ) = \frac{3}{2}

sin (6 π)

cos (2 5^{\circ})

csc (\frac{4 π}{3})

cos (3 θ) = 1