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人気のある三角関数 >

4cos^2(2x)-2=0

解

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

解

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

+1

度

x = 22. 5^{\circ} + 18 0^{\circ} n, x = 157. 5^{\circ} + 18 0^{\circ} n, x = 67. 5^{\circ} + 18 0^{\circ} n, x = - 67. 5^{\circ} + 18 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (2 x) = u

4 u^{2} - 2 = 0

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

4 u^{2} - 2 = 0

2

を右側に移動します

4 u^{2} - 2 = 0

両辺に

2

を足す

4 u^{2} - 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

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 (2 x)

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

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

cos (2 x) = \frac{1}{2} : x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

解く

2 x = arccos (\frac{1}{2}) + 2 πn : x = \frac{π}{8} + πn

2 x = 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

2 x = \frac{π}{4} + 2 πn

以下で両辺を割る

2

2 x = \frac{π}{4} + 2 πn

以下で両辺を割る

2

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

簡素化

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

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

分数の規則を適用する:

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

= \frac{π}{4 \cdot 2}

数を乗じる:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

x = \frac{π}{8} + πn

x = \frac{π}{8} + πn

x = \frac{π}{8} + πn

解く

2 x = 2 π - arccos (\frac{1}{2}) + 2 πn : x = π - \frac{π}{8} + πn

2 x = 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

2 x = 2 π - \frac{π}{4} + 2 πn

以下で両辺を割る

2

2 x = 2 π - \frac{π}{4} + 2 πn

以下で両辺を割る

2

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

簡素化

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

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

\frac{2 π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : π - \frac{π}{8} + πn

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

\frac{2 π}{2} = π

\frac{2 π}{2}

数を割る:

\frac{2}{2} = 1

= π

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

分数の規則を適用する:

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

= \frac{π}{4 \cdot 2}

数を乗じる:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

= π - \frac{π}{8} + πn

x = π - \frac{π}{8} + πn

x = π - \frac{π}{8} + πn

x = π - \frac{π}{8} + πn

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn

cos (2 x) = - \frac{1}{2} : x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

解く

2 x = arccos (- \frac{1}{2}) + 2 πn : x = \frac{3 π}{8} + πn

2 x = 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

2 x = \frac{3 π}{4} + 2 πn

以下で両辺を割る

2

2 x = \frac{3 π}{4} + 2 πn

以下で両辺を割る

2

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

簡素化

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

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

\frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2} : \frac{3 π}{8} + πn

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

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

分数の規則を適用する:

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

= \frac{3 π}{4 \cdot 2}

数を乗じる:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

解く

2 x = - arccos (- \frac{1}{2}) + 2 πn : x = - \frac{3 π}{8} + πn

2 x = - 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

2 x = - \frac{3 π}{4} + 2 πn

以下で両辺を割る

2

2 x = - \frac{3 π}{4} + 2 πn

以下で両辺を割る

2

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

簡素化

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

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

- \frac{\frac{3 π}{4}}{2} + \frac{2 πn}{2} : - \frac{3 π}{8} + πn

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

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

分数の規則を適用する:

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

= \frac{3 π}{4 \cdot 2}

数を乗じる:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

= - \frac{3 π}{8} + πn

x = - \frac{3 π}{8} + πn

x = - \frac{3 π}{8} + πn

x = - \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

すべての解を組み合わせる

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

グラフ

人気の例

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tan (a) = \frac{20}{21}

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

3tan(θ)sin(θ)-2tan(θ)=0

3 tan (θ) sin (θ) - 2 tan (θ) = 0

-8csc^2(x)+2cot(x)=-23

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