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人気のある三角関数 >

2sec^2(a)+tan^2(a)=3

解

2 sec^{2} (a) + tan^{2} (a) = 3

解

a = \frac{π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn, a = \frac{7 π}{6} + 2 πn

+1

度

a = 3 0^{\circ} + 36 0^{\circ} n, a = 33 0^{\circ} + 36 0^{\circ} n, a = 15 0^{\circ} + 36 0^{\circ} n, a = 21 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 sec^{2} (a) + tan^{2} (a) = 3

両辺から

3

を引く

2 sec^{2} (a) + tan^{2} (a) - 3 = 0

三角関数の公式を使用して書き換える

- 3 + tan^{2} (a) + 2 sec^{2} (a)

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

sec^{2} (a) + 2 sec^{2} (a) = 3 sec^{2} (a)

= 3 sec^{2} (a) - 3 - 1

数を引く:

- 3 - 1 = - 4

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

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

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

置換で解く

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

仮定:

sec (a) = u

- 4 + 3 u^{2} = 0

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

- 4 + 3 u^{2} = 0

4

を右側に移動します

- 4 + 3 u^{2} = 0

両辺に

4

を足す

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

簡素化

3 u^{2} = 4

3 u^{2} = 4

以下で両辺を割る

3

3 u^{2} = 4

以下で両辺を割る

3

\frac{3 u ^{2}}{3} = \frac{4}{3}

簡素化

u^{2} = \frac{4}{3}

u^{2} = \frac{4}{3}

x^{2} = f (a)

の場合, 解は

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

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

\frac{4}{3} = \frac{2 3}{3}

\frac{4}{3}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{4}{3}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{3}

有理化する

\frac{2}{3} : \frac{2 3}{3}

\frac{2}{3}

共役で乗じる

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

累乗根の規則を適用する:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

= \frac{2 3}{3}

- \frac{4}{3} = - \frac{2 3}{3}

- \frac{4}{3}

簡素化

\frac{4}{3} : \frac{2}{3}

\frac{4}{3}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{4}{3}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2}{3}

= - \frac{2}{3}

有理化する

- \frac{2}{3} : - \frac{2 3}{3}

- \frac{2}{3}

共役で乗じる

\frac{3}{3}

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

33 = 3

33

累乗根の規則を適用する:

a a = a

33 = 3

= 3

= - \frac{2 3}{3}

= - \frac{2 3}{3}

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

代用を戻す

u = sec (a)

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

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

sec (a) = \frac{2 3}{3} : a = \frac{π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn

sec (a) = \frac{2 3}{3}

以下の一般解

sec (a) = \frac{2 3}{3}

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}

a = \frac{π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn

a = \frac{π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn

sec (a) = - \frac{2 3}{3} : a = \frac{5 π}{6} + 2 πn, a = \frac{7 π}{6} + 2 πn

sec (a) = - \frac{2 3}{3}

以下の一般解

sec (a) = - \frac{2 3}{3}

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}

a = \frac{5 π}{6} + 2 πn, a = \frac{7 π}{6} + 2 πn

a = \frac{5 π}{6} + 2 πn, a = \frac{7 π}{6} + 2 πn

すべての解を組み合わせる

a = \frac{π}{6} + 2 πn, a = \frac{11 π}{6} + 2 πn, a = \frac{5 π}{6} + 2 πn, a = \frac{7 π}{6} + 2 πn

グラフ

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