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人気のある三角関数 >

arccos(x)+arccos(2x)= pi/2

解

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

解

x = \frac{1}{5}

解答ステップ

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

三角関数の公式を使用して書き換える

arccos (x) + arccos (2 x)

和・積の公式を使用する:

arccos (s) + arccos (t) = arccos (s t - (1 - s^{2}) (1 - t^{2}))

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

arccos (x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2})) = \frac{π}{2}

三角関数の逆数プロパティを適用する

arccos (x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2})) = \frac{π}{2}

arccos (x) = a \Rightarrow x = cos (a)

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

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

次の自明恒等式を使用する:

cos (\frac{π}{2}) = 0

cos (\frac{π}{2})

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

= 0

= 0

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

解く

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0 : x = \frac{1}{5}, x = - \frac{1}{5}

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

拡張

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) : 2 x^{2} - 1 - 5 x^{2} + 4 x^{4}

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

x \cdot 2 x = 2 x^{2}

x \cdot 2 x

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

xx = x^{1 + 1}

= 2 x^{1 + 1}

数を足す:

1 + 1 = 2

= 2 x^{2}

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

拡張

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

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

(1 - x^{2}) (1 - (2 x)^{2}) = 1 - 5 x^{2} + 4 x^{4}

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

(2 x)^{2} = 4 x^{2}

(2 x)^{2}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} x^{2}

2^{2} = 4

= 4 x^{2}

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

拡張

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

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

FOIL メソッドを適用する:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = - x^{2}, c = 1, d = - 4 x^{2}

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

マイナス・プラスの規則を適用する

+ (- a) = - a, (- a) (- b) = ab

= 1 \cdot 1 - 1 \cdot 4 x^{2} - 1 \cdot x^{2} + 4 x^{2} x^{2}

簡素化

1 \cdot 1 - 1 \cdot 4 x^{2} - 1 \cdot x^{2} + 4 x^{2} x^{2} : 1 - 5 x^{2} + 4 x^{4}

1 \cdot 1 - 1 \cdot 4 x^{2} - 1 \cdot x^{2} + 4 x^{2} x^{2}

1 \cdot 1 = 1

1 \cdot 1

数を乗じる:

1 \cdot 1 = 1

= 1

1 \cdot 4 x^{2} = 4 x^{2}

1 \cdot 4 x^{2}

数を乗じる:

1 \cdot 4 = 4

= 4 x^{2}

1 \cdot x^{2} = x^{2}

1 \cdot x^{2}

乗算:

1 \cdot x^{2} = x^{2}

= x^{2}

4 x^{2} x^{2} = 4 x^{4}

4 x^{2} x^{2}

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

x^{2} x^{2} = x^{2 + 2}

= 4 x^{2 + 2}

数を足す:

2 + 2 = 4

= 4 x^{4}

= 1 - 4 x^{2} - x^{2} + 4 x^{4}

類似した元を足す:

- 4 x^{2} - x^{2} = - 5 x^{2}

= 1 - 5 x^{2} + 4 x^{4}

= 1 - 5 x^{2} + 4 x^{4}

= 1 - 5 x^{2} + 4 x^{4}

= 2 x^{2} - 4 x^{4} - 5 x^{2} + 1

= 2 x^{2} - 1 - 5 x^{2} + 4 x^{4}

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} = 0

平方根を削除する

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} = 0

両辺から

2 x^{2}

を引く

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} - 2 x^{2} = 0 - 2 x^{2}

簡素化

- 1 - 5 x^{2} + 4 x^{4} = - 2 x^{2}

両辺を2乗する:

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

2 x^{2} - 1 - 5 x^{2} + 4 x^{4} = 0

(- 1 - 5 x^{2} + 4 x^{4})^{2} = (- 2 x^{2})^{2}

拡張

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

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

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 1 - 5 x^{2} + 4 x^{4})^{2} = (1 - 5 x^{2} + 4 x^{4})^{2}

= (1 - 5 x^{2} + 4 x^{4})^{2}

累乗根の規則を適用する:

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

= ((1 - 5 x^{2} + 4 x^{4})^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= (1 - 5 x^{2} + 4 x^{4})^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分数を乗じる:

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

= \frac{1 \cdot 2}{2}

共通因数を約分する:

2

= 1

= 1 - 5 x^{2} + 4 x^{4}

拡張

(- 2 x^{2})^{2} : 4 x^{4}

(- 2 x^{2})^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

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

= (2 x^{2})^{2}

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} (x^{2})^{2}

(x^{2})^{2} : x^{4}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= x^{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= x^{4}

= 2^{2} x^{4}

2^{2} = 4

= 4 x^{4}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

解く

1 - 5 x^{2} + 4 x^{4} = 4 x^{4} : x = \frac{1}{5}, x = - \frac{1}{5}

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

1

を右側に移動します

1 - 5 x^{2} + 4 x^{4} = 4 x^{4}

両辺から

1

を引く

1 - 5 x^{2} + 4 x^{4} - 1 = 4 x^{4} - 1

簡素化

- 5 x^{2} + 4 x^{4} = 4 x^{4} - 1

- 5 x^{2} + 4 x^{4} = 4 x^{4} - 1

4 x^{4}

を左側に移動します

- 5 x^{2} + 4 x^{4} = 4 x^{4} - 1

両辺から

4 x^{4}

を引く

- 5 x^{2} + 4 x^{4} - 4 x^{4} = 4 x^{4} - 1 - 4 x^{4}

簡素化

- 5 x^{2} = - 1

- 5 x^{2} = - 1

以下で両辺を割る

- 5

- 5 x^{2} = - 1

以下で両辺を割る

- 5

\frac{- 5 x ^{2}}{- 5} = \frac{- 1}{- 5}

簡素化

x^{2} = \frac{1}{5}

x^{2} = \frac{1}{5}

x^{2} = f (a)

の場合, 解は

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

x = \frac{1}{5}, x = - \frac{1}{5}

\frac{1}{5} = \frac{1}{5}

\frac{1}{5}

累乗根の規則を適用する:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= \frac{1}{5}

累乗根の規則を適用する:

1 = 1

1 = 1

= \frac{1}{5}

- \frac{1}{5} = - \frac{1}{5}

- \frac{1}{5}

累乗根の規則を適用する:

\frac{a}{b} = \frac{a}{b}, a \geq 0, b \geq 0

= - \frac{1}{5}

累乗根の規則を適用する:

1 = 1

1 = 1

= - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

解を検算する:

x = \frac{1}{5}

真

, x = - \frac{1}{5}

真

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

に当てはめて解を確認する

equationに一致しないものを削除する。

挿入

x = \frac{1}{5} :

真

(\frac{1}{5}) \cdot 2 (\frac{1}{5}) - (1 - (\frac{1}{5})^{2}) (1 - (2 (\frac{1}{5}))^{2}) = 0

(\frac{1}{5}) \cdot 2 (\frac{1}{5}) - (1 - (\frac{1}{5})^{2}) (1 - (2 (\frac{1}{5}))^{2}) = 0

(\frac{1}{5}) \cdot 2 (\frac{1}{5}) - (1 - (\frac{1}{5})^{2}) (1 - (2 (\frac{1}{5}))^{2})

括弧を削除する:

(a) = a

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

\frac{1}{5} \cdot 2 \cdot \frac{1}{5} = \frac{2}{5}

\frac{1}{5} \cdot 2 \cdot \frac{1}{5}

分数を乗じる:

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

= \frac{1 \cdot 1 \cdot 2}{5 5}

数を乗じる:

1 \cdot 1 \cdot 2 = 2

= \frac{2}{5 5}

55 = 5

55

累乗根の規則を適用する:

a a = a

55 = 5

= 5

= \frac{2}{5}

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

(1 - (\frac{1}{5})^{2}) (1 - (2 \cdot \frac{1}{5})^{2})

(\frac{1}{5})^{2} = \frac{1}{5}

(\frac{1}{5})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

累乗根の規則を適用する:

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

= (5^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分数を乗じる:

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

= \frac{1 \cdot 2}{2}

共通因数を約分する:

2

= 1

= 5

= \frac{1 ^{2}}{5}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{5}

= (- \frac{1}{5} + 1) (- (2 \cdot \frac{1}{5})^{2} + 1)

(2 \cdot \frac{1}{5})^{2} = \frac{4}{5}

(2 \cdot \frac{1}{5})^{2}

乗じる

2 \cdot \frac{1}{5} : \frac{2}{5}

2 \cdot \frac{1}{5}

分数を乗じる:

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

= \frac{1 \cdot 2}{5}

数を乗じる:

1 \cdot 2 = 2

= \frac{2}{5}

= (\frac{2}{5})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{2 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

累乗根の規則を適用する:

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

= (5^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分数を乗じる:

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

= \frac{1 \cdot 2}{2}

共通因数を約分する:

2

= 1

= 5

= \frac{2 ^{2}}{5}

2^{2} = 4

= \frac{4}{5}

= (- \frac{1}{5} + 1) (- \frac{4}{5} + 1)

結合

1 - \frac{1}{5} : \frac{4}{5}

1 - \frac{1}{5}

元を分数に変換する:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{1}{5}

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{1 \cdot 5 - 1}{5}

1 \cdot 5 - 1 = 4

1 \cdot 5 - 1

数を乗じる:

1 \cdot 5 = 5

= 5 - 1

数を引く:

5 - 1 = 4

= 4

= \frac{4}{5}

= \frac{4}{5} (- \frac{4}{5} + 1)

結合

1 - \frac{4}{5} : \frac{1}{5}

1 - \frac{4}{5}

元を分数に変換する:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{4}{5}

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{1 \cdot 5 - 4}{5}

1 \cdot 5 - 4 = 1

1 \cdot 5 - 4

数を乗じる:

1 \cdot 5 = 5

= 5 - 4

数を引く:

5 - 4 = 1

= 1

= \frac{1}{5}

= \frac{4}{5} \cdot \frac{1}{5}

乗じる

\frac{4}{5} \cdot \frac{1}{5} : \frac{4}{25}

\frac{4}{5} \cdot \frac{1}{5}

分数を乗じる:

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

= \frac{4 \cdot 1}{5 \cdot 5}

数を乗じる:

4 \cdot 1 = 4

= \frac{4}{5 \cdot 5}

数を乗じる:

5 \cdot 5 = 25

= \frac{4}{25}

= \frac{4}{25}

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}{25}

25 = 5

25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

= \frac{4}{5}

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}{5}

= \frac{2}{5} - \frac{2}{5}

類似した元を足す:

\frac{2}{5} - \frac{2}{5} = 0

= 0

0 = 0

真

挿入

x = - \frac{1}{5} :

真

(- \frac{1}{5}) \cdot 2 (- \frac{1}{5}) - (1 - (- \frac{1}{5})^{2}) (1 - (2 (- \frac{1}{5}))^{2}) = 0

(- \frac{1}{5}) \cdot 2 (- \frac{1}{5}) - (1 - (- \frac{1}{5})^{2}) (1 - (2 (- \frac{1}{5}))^{2}) = 0

(- \frac{1}{5}) \cdot 2 (- \frac{1}{5}) - (1 - (- \frac{1}{5})^{2}) (1 - (2 (- \frac{1}{5}))^{2})

括弧を削除する:

(- a) = - a, - (- a) = a

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

\frac{1}{5} \cdot 2 \cdot \frac{1}{5} = \frac{2}{5}

\frac{1}{5} \cdot 2 \cdot \frac{1}{5}

分数を乗じる:

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

= \frac{1 \cdot 1 \cdot 2}{5 5}

数を乗じる:

1 \cdot 1 \cdot 2 = 2

= \frac{2}{5 5}

55 = 5

55

累乗根の規則を適用する:

a a = a

55 = 5

= 5

= \frac{2}{5}

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

(1 - (- \frac{1}{5})^{2}) (1 - (- 2 \cdot \frac{1}{5})^{2})

(- \frac{1}{5})^{2} = \frac{1}{5}

(- \frac{1}{5})^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- \frac{1}{5})^{2} = (\frac{1}{5})^{2}

= (\frac{1}{5})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

累乗根の規則を適用する:

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

= (5^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分数を乗じる:

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

= \frac{1 \cdot 2}{2}

共通因数を約分する:

2

= 1

= 5

= \frac{1 ^{2}}{5}

規則を適用

1^{a} = 1

1^{2} = 1

= \frac{1}{5}

= (- \frac{1}{5} + 1) (- (- 2 \cdot \frac{1}{5})^{2} + 1)

(- 2 \cdot \frac{1}{5})^{2} = \frac{4}{5}

(- 2 \cdot \frac{1}{5})^{2}

乗じる

- 2 \cdot \frac{1}{5} : - \frac{2}{5}

- 2 \cdot \frac{1}{5}

分数を乗じる:

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

= - \frac{1 \cdot 2}{5}

数を乗じる:

1 \cdot 2 = 2

= - \frac{2}{5}

= (- \frac{2}{5})^{2}

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- \frac{2}{5})^{2} = (\frac{2}{5})^{2}

= (\frac{2}{5})^{2}

指数の規則を適用する:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{2 ^{2}}{( 5 ) ^{2}}

(5)^{2} : 5

累乗根の規則を適用する:

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

= (5^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分数を乗じる:

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

= \frac{1 \cdot 2}{2}

共通因数を約分する:

2

= 1

= 5

= \frac{2 ^{2}}{5}

2^{2} = 4

= \frac{4}{5}

= (- \frac{1}{5} + 1) (- \frac{4}{5} + 1)

結合

1 - \frac{1}{5} : \frac{4}{5}

1 - \frac{1}{5}

元を分数に変換する:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{1}{5}

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{1 \cdot 5 - 1}{5}

1 \cdot 5 - 1 = 4

1 \cdot 5 - 1

数を乗じる:

1 \cdot 5 = 5

= 5 - 1

数を引く:

5 - 1 = 4

= 4

= \frac{4}{5}

= \frac{4}{5} (- \frac{4}{5} + 1)

結合

1 - \frac{4}{5} : \frac{1}{5}

1 - \frac{4}{5}

元を分数に変換する:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{4}{5}

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{1 \cdot 5 - 4}{5}

1 \cdot 5 - 4 = 1

1 \cdot 5 - 4

数を乗じる:

1 \cdot 5 = 5

= 5 - 4

数を引く:

5 - 4 = 1

= 1

= \frac{1}{5}

= \frac{4}{5} \cdot \frac{1}{5}

乗じる

\frac{4}{5} \cdot \frac{1}{5} : \frac{4}{25}

\frac{4}{5} \cdot \frac{1}{5}

分数を乗じる:

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

= \frac{4 \cdot 1}{5 \cdot 5}

数を乗じる:

4 \cdot 1 = 4

= \frac{4}{5 \cdot 5}

数を乗じる:

5 \cdot 5 = 25

= \frac{4}{25}

= \frac{4}{25}

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}{25}

25 = 5

25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

= \frac{4}{5}

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}{5}

= \frac{2}{5} - \frac{2}{5}

類似した元を足す:

\frac{2}{5} - \frac{2}{5} = 0

= 0

0 = 0

真

解答は

x = \frac{1}{5}, x = - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

元のequationに当てはめて解を検算する

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

に当てはめて解を確認する

equationに一致しないものを削除する。

解答を確認する

\frac{1}{5} :

真

\frac{1}{5}

挿入

n = 1

\frac{1}{5}

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

の挿入向け

x = \frac{1}{5}

arccos (\frac{1}{5}) + arccos (2 \cdot \frac{1}{5}) = \frac{π}{2}

改良

1.57079 \dots = 1.57079 \dots

\Rightarrow 真

解答を確認する

- \frac{1}{5} :

偽

- \frac{1}{5}

挿入

n = 1

- \frac{1}{5}

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

の挿入向け

x = - \frac{1}{5}

arccos (- \frac{1}{5}) + arccos (2 (- \frac{1}{5})) = \frac{π}{2}

改良

4.71238 \dots = 1.57079 \dots

\Rightarrow 偽

x = \frac{1}{5}

グラフ

人気の例

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

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

-2-3sin(x)=1-6sin(x)

- 2 - 3 sin (x) = 1 - 6 sin (x)

cos (x) = - 0.2

3cos^2(x)-2cos(x)-1=0

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

2sin(x)-cos(2x)=0

2 sin (x) - cos (2 x) = 0