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人気のある三角関数 >

5sec^2(x)-5=0,0<= x<2pi

解

5 sec^{2} (x) - 5 = 0, 0 \leq x < 2 π

解

x = 0, x = π

+1

度

x = 0^{\circ}, x = 18 0^{\circ}

解答ステップ

5 sec^{2} (x) - 5 = 0, 0 \leq x < 2 π

置換で解く

5 sec^{2} (x) - 5 = 0

仮定:

sec (x) = u

5 u^{2} - 5 = 0

5 u^{2} - 5 = 0 : u = 1, u = - 1

5 u^{2} - 5 = 0

5

を右側に移動します

5 u^{2} - 5 = 0

両辺に

5

を足す

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

簡素化

5 u^{2} = 5

5 u^{2} = 5

以下で両辺を割る

5

5 u^{2} = 5

以下で両辺を割る

5

\frac{5 u ^{2}}{5} = \frac{5}{5}

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, 0 \leq x < 2 π : x = 0

sec (x) = 1, 0 \leq x < 2 π

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

範囲の解答

0 \leq x < 2 π

x = 0

sec (x) = - 1, 0 \leq x < 2 π : x = π

sec (x) = - 1, 0 \leq x < 2 π

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

範囲の解答

0 \leq x < 2 π

x = π

すべての解を組み合わせる

x = 0, x = π

グラフ

人気の例

2 sin (5 x) + 3 = 2

5sin(y)-5=3sin(y)-4

5 sin (y) - 5 = 3 sin (y) - 4

2cos^2(x)+3cos(x)-3=0

2 cos^{2} (x) + 3 cos (x) - 3 = 0

3sin(x)-5cos(x)=0

3 sin (x) - 5 cos (x) = 0

sin^{2} (x) = 0.85