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人気のある三角関数 >

sec^2(x)-1=0

解

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

解

x = 2 πn, x = π + 2 πn

+1

度

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

解答ステップ

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

置換で解く

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

仮定:

sec (x) = u

u^{2} - 1 = 0

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

u^{2} - 1 = 0

1

を右側に移動します

u^{2} - 1 = 0

両辺に

1

を足す

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

簡素化

u^{2} = 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

代用を戻す

u = sec (x)

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

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

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) = - 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 = 2 πn, x = π + 2 πn

グラフ

人気の例

tan (θ) = \frac{3}{4}

4 cos^{2} (x) = 1

2sin(x)cos(x)-sin(x)=0

2 sin (x) cos (x) - sin (x) = 0

cos (x) = \frac{3}{5}

3sec^2(x)-2tan^2(x)-4=0

3 sec^{2} (x) - 2 tan^{2} (x) - 4 = 0