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人気のある三角関数 >

sin^2(x)+5cos^2(x)=3

解

sin^{2} (x) + 5 cos^{2} (x) = 3

解

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

+1

度

x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n, x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

解答ステップ

sin^{2} (x) + 5 cos^{2} (x) = 3

両辺から

3

を引く

sin^{2} (x) + 5 cos^{2} (x) - 3 = 0

三角関数の公式を使用して書き換える

- 3 + sin^{2} (x) + 5 cos^{2} (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 3 + sin^{2} (x) + 5 (1 - sin^{2} (x))

簡素化

- 3 + sin^{2} (x) + 5 (1 - sin^{2} (x)) : - 4 sin^{2} (x) + 2

- 3 + sin^{2} (x) + 5 (1 - sin^{2} (x))

拡張

5 (1 - sin^{2} (x)) : 5 - 5 sin^{2} (x)

5 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 5, b = 1, c = sin^{2} (x)

= 5 \cdot 1 - 5 sin^{2} (x)

数を乗じる:

5 \cdot 1 = 5

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

= - 3 + sin^{2} (x) + 5 - 5 sin^{2} (x)

簡素化

- 3 + sin^{2} (x) + 5 - 5 sin^{2} (x) : - 4 sin^{2} (x) + 2

- 3 + sin^{2} (x) + 5 - 5 sin^{2} (x)

条件のようなグループ

= sin^{2} (x) - 5 sin^{2} (x) - 3 + 5

類似した元を足す:

sin^{2} (x) - 5 sin^{2} (x) = - 4 sin^{2} (x)

= - 4 sin^{2} (x) - 3 + 5

数を足す/引く:

- 3 + 5 = 2

= - 4 sin^{2} (x) + 2

= - 4 sin^{2} (x) + 2

= - 4 sin^{2} (x) + 2

2 - 4 sin^{2} (x) = 0

置換で解く

2 - 4 sin^{2} (x) = 0

仮定:

sin (x) = u

2 - 4 u^{2} = 0

2 - 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

2 - 4 u^{2} = 0

2

を右側に移動します

2 - 4 u^{2} = 0

両辺から

2

を引く

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

簡素化

- 4 u^{2} = - 2

- 4 u^{2} = - 2

以下で両辺を割る

- 4

- 4 u^{2} = - 2

以下で両辺を割る

- 4

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

簡素化

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

簡素化

\frac{- 4 u ^{2}}{- 4} : u^{2}

\frac{- 4 u ^{2}}{- 4}

分数の規則を適用する:

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

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

数を割る:

\frac{4}{4} = 1

= u^{2}

簡素化

\frac{- 2}{- 4} : \frac{1}{2}

\frac{- 2}{- 4}

分数の規則を適用する:

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

x^{2} = f (a)

の場合, 解は

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

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

代用を戻す

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = \frac{1}{2}

以下の一般解

sin (x) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

sin (x) = - \frac{1}{2}

以下の一般解

sin (x) = - \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

すべての解を組み合わせる

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

グラフ

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