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人気のある三角関数 >

2cos^2(x/2)=1

解

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

解

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

+1

度

x = 9 0^{\circ} + 72 0^{\circ} n, x = 63 0^{\circ} + 72 0^{\circ} n, x = 27 0^{\circ} + 72 0^{\circ} n, x = - 27 0^{\circ} + 72 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (\frac{x}{2}) = u

2 u^{2} = 1

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

2 u^{2} = 1

以下で両辺を割る

2

2 u^{2} = 1

以下で両辺を割る

2

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

簡素化

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 = cos (\frac{x}{2})

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

\frac{x}{2} = arccos (\frac{1}{2}) + 2 πn, \frac{x}{2} = 2 π - arccos (\frac{1}{2}) + 2 πn

\frac{x}{2} = arccos (\frac{1}{2}) + 2 πn, \frac{x}{2} = 2 π - arccos (\frac{1}{2}) + 2 πn

解く

\frac{x}{2} = arccos (\frac{1}{2}) + 2 πn : x = \frac{π}{2} + 4 πn

\frac{x}{2} = arccos (\frac{1}{2}) + 2 πn

簡素化

arccos (\frac{1}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{1}{2}) + 2 πn

次の自明恒等式を使用する:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{4} + 2 πn

\frac{x}{2} = \frac{π}{4} + 2 πn

以下で両辺を乗じる:

2

\frac{x}{2} = \frac{π}{4} + 2 πn

以下で両辺を乗じる:

2

\frac{2 x}{2} = 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} = 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

2 \cdot \frac{π}{4} + 2 \cdot 2 πn : \frac{π}{2} + 4 πn

2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分数を乗じる:

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

= \frac{π 2}{4}

共通因数を約分する:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

x = \frac{π}{2} + 4 πn

解く

\frac{x}{2} = 2 π - arccos (\frac{1}{2}) + 2 πn : x = 4 π - \frac{π}{2} + 4 πn

\frac{x}{2} = 2 π - arccos (\frac{1}{2}) + 2 πn

簡素化

2 π - arccos (\frac{1}{2}) + 2 πn : 2 π - \frac{π}{4} + 2 πn

2 π - arccos (\frac{1}{2}) + 2 πn

次の自明恒等式を使用する:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{4} + 2 πn

\frac{x}{2} = 2 π - \frac{π}{4} + 2 πn

以下で両辺を乗じる:

2

\frac{x}{2} = 2 π - \frac{π}{4} + 2 πn

以下で両辺を乗じる:

2

\frac{2 x}{2} = 2 \cdot 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} = 2 \cdot 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

2 \cdot 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn : 4 π - \frac{π}{2} + 4 πn

2 \cdot 2 π - 2 \cdot \frac{π}{4} + 2 \cdot 2 πn

2 \cdot 2 π = 4 π

2 \cdot 2 π

数を乗じる:

2 \cdot 2 = 4

= 4 π

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

分数を乗じる:

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

= \frac{π 2}{4}

共通因数を約分する:

2

= \frac{π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 4 πn

= 4 π - \frac{π}{2} + 4 πn

x = 4 π - \frac{π}{2} + 4 πn

x = 4 π - \frac{π}{2} + 4 πn

x = 4 π - \frac{π}{2} + 4 πn

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

\frac{x}{2} = arccos (- \frac{1}{2}) + 2 πn, \frac{x}{2} = - arccos (- \frac{1}{2}) + 2 πn

\frac{x}{2} = arccos (- \frac{1}{2}) + 2 πn, \frac{x}{2} = - arccos (- \frac{1}{2}) + 2 πn

解く

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

\frac{x}{2} = arccos (- \frac{1}{2}) + 2 πn

簡素化

arccos (- \frac{1}{2}) + 2 πn : \frac{3 π}{4} + 2 πn

arccos (- \frac{1}{2}) + 2 πn

次の自明恒等式を使用する:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{3 π}{4} + 2 πn

\frac{x}{2} = \frac{3 π}{4} + 2 πn

以下で両辺を乗じる:

2

\frac{x}{2} = \frac{3 π}{4} + 2 πn

以下で両辺を乗じる:

2

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

簡素化

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

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn : \frac{3 π}{2} + 4 πn

2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{4} = \frac{3 π}{2}

2 \cdot \frac{3 π}{4}

分数を乗じる:

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

= \frac{3 π 2}{4}

数を乗じる:

3 \cdot 2 = 6

= \frac{6 π}{4}

共通因数を約分する:

2

= \frac{3 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 4 πn

= \frac{3 π}{2} + 4 πn

x = \frac{3 π}{2} + 4 πn

x = \frac{3 π}{2} + 4 πn

x = \frac{3 π}{2} + 4 πn

解く

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

\frac{x}{2} = - arccos (- \frac{1}{2}) + 2 πn

簡素化

- arccos (- \frac{1}{2}) + 2 πn : - \frac{3 π}{4} + 2 πn

- arccos (- \frac{1}{2}) + 2 πn

次の自明恒等式を使用する:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{3 π}{4} + 2 πn

\frac{x}{2} = - \frac{3 π}{4} + 2 πn

以下で両辺を乗じる:

2

\frac{x}{2} = - \frac{3 π}{4} + 2 πn

以下で両辺を乗じる:

2

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

簡素化

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

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

- 2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn : - \frac{3 π}{2} + 4 πn

- 2 \cdot \frac{3 π}{4} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{4} = \frac{3 π}{2}

2 \cdot \frac{3 π}{4}

分数を乗じる:

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

= \frac{3 π 2}{4}

数を乗じる:

3 \cdot 2 = 6

= \frac{6 π}{4}

共通因数を約分する:

2

= \frac{3 π}{2}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 4 πn

= - \frac{3 π}{2} + 4 πn

x = - \frac{3 π}{2} + 4 πn

x = - \frac{3 π}{2} + 4 πn

x = - \frac{3 π}{2} + 4 πn

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

すべての解を組み合わせる

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

グラフ

人気の例

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