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人気のある三角関数 >

cos(2x)-sin^2(x)= 1/4

解

cos (2 x) - sin^{2} (x) = \frac{1}{4}

解

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

+1

度

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

解答ステップ

cos (2 x) - sin^{2} (x) = \frac{1}{4}

両辺から

\frac{1}{4}

を引く

cos (2 x) - sin^{2} (x) - \frac{1}{4} = 0

簡素化

cos (2 x) - sin^{2} (x) - \frac{1}{4} : \frac{4 cos ( 2 x ) - 4 sin ^{2} ( x ) - 1}{4}

cos (2 x) - sin^{2} (x) - \frac{1}{4}

元を分数に変換する:

cos (2 x) = \frac{cos ( 2 x ) 4}{4}, sin^{2} (x) = \frac{sin ^{2} ( x ) 4}{4}

= \frac{cos ( 2 x ) \cdot 4}{4} - \frac{sin ^{2} ( x ) \cdot 4}{4} - \frac{1}{4}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ( 2 x ) \cdot 4 - sin ^{2} ( x ) \cdot 4 - 1}{4}

\frac{4 cos ( 2 x ) - 4 sin ^{2} ( x ) - 1}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 cos (2 x) - 4 sin^{2} (x) - 1 = 0

三角関数の公式を使用して書き換える

- 1 + 4 cos (2 x) - 4 sin^{2} (x)

2倍角の公式を使用:

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

= - 1 + 4 (1 - 2 sin^{2} (x)) - 4 sin^{2} (x)

簡素化

- 1 + 4 (1 - 2 sin^{2} (x)) - 4 sin^{2} (x) : - 12 sin^{2} (x) + 3

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

拡張

4 (1 - 2 sin^{2} (x)) : 4 - 8 sin^{2} (x)

4 (1 - 2 sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 1, c = 2 sin^{2} (x)

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

簡素化

4 \cdot 1 - 4 \cdot 2 sin^{2} (x) : 4 - 8 sin^{2} (x)

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

数を乗じる:

4 \cdot 1 = 4

= 4 - 4 \cdot 2 sin^{2} (x)

数を乗じる:

4 \cdot 2 = 8

= 4 - 8 sin^{2} (x)

= 4 - 8 sin^{2} (x)

= - 1 + 4 - 8 sin^{2} (x) - 4 sin^{2} (x)

簡素化

- 1 + 4 - 8 sin^{2} (x) - 4 sin^{2} (x) : - 12 sin^{2} (x) + 3

- 1 + 4 - 8 sin^{2} (x) - 4 sin^{2} (x)

類似した元を足す:

- 8 sin^{2} (x) - 4 sin^{2} (x) = - 12 sin^{2} (x)

= - 1 + 4 - 12 sin^{2} (x)

数を足す/引く:

- 1 + 4 = 3

= - 12 sin^{2} (x) + 3

= - 12 sin^{2} (x) + 3

= - 12 sin^{2} (x) + 3

3 - 12 sin^{2} (x) = 0

置換で解く

3 - 12 sin^{2} (x) = 0

仮定:

sin (x) = u

3 - 12 u^{2} = 0

3 - 12 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

3 - 12 u^{2} = 0

3

を右側に移動します

3 - 12 u^{2} = 0

両辺から

3

を引く

3 - 12 u^{2} - 3 = 0 - 3

簡素化

- 12 u^{2} = - 3

- 12 u^{2} = - 3

以下で両辺を割る

- 12

- 12 u^{2} = - 3

以下で両辺を割る

- 12

\frac{- 12 u ^{2}}{- 12} = \frac{- 3}{- 12}

簡素化

\frac{- 12 u ^{2}}{- 12} = \frac{- 3}{- 12}

簡素化

\frac{- 12 u ^{2}}{- 12} : u^{2}

\frac{- 12 u ^{2}}{- 12}

分数の規則を適用する:

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

= \frac{12 u ^{2}}{12}

数を割る:

\frac{12}{12} = 1

= u^{2}

簡素化

\frac{- 3}{- 12} : \frac{1}{4}

\frac{- 3}{- 12}

分数の規則を適用する:

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

= \frac{3}{12}

共通因数を約分する:

3

= \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{4}, u = - \frac{1}{4}

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

\frac{1}{4}

累乗根の規則を適用する:

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{4}

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:

2^{2} = 2

= 2

= \frac{1}{2}

規則を適用

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

簡素化

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

\frac{1}{4}

累乗根の規則を適用する:

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{4}

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:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

規則を適用

1 = 1

= - \frac{1}{2}

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

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

すべての解を組み合わせる

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

グラフ

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