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人気のある三角関数 >

(cos^2(x))/(1-cos(x))=1+cos(x)

解

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

解

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

+1

度

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

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

以下で両辺を乗じる:

1 - u

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

以下で両辺を乗じる:

1 - u

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

簡素化

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

簡素化

\frac{u ^{2}}{1 - u} (1 - u) : u^{2}

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

分数を乗じる:

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

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

共通因数を約分する:

1 - u

= u^{2}

簡素化

1 \cdot (1 - u) : 1 - u

1 \cdot (1 - u)

乗算:

1 \cdot (1 - u) = (1 - u)

= (1 - u)

括弧を削除する:

(a) = a

= 1 - u

u^{2} = 1 - u + u (1 - u)

u^{2} = 1 - u + u (1 - u)

u^{2} = 1 - u + u (1 - u)

解く

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

u^{2} = 1 - u + u (1 - u)

拡張

1 - u + u (1 - u) : 1 - u^{2}

1 - u + u (1 - u)

拡張

u (1 - u) : u - u^{2}

u (1 - u)

分配法則を適用する:

a (b - c) = ab - a c

a = u, b = 1, c = u

= u \cdot 1 - uu

= 1 \cdot u - uu

簡素化

1 \cdot u - uu : u - u^{2}

1 \cdot u - uu

1 \cdot u = u

1 \cdot u

乗算:

1 \cdot u = u

= u

uu = u^{2}

uu

指数の規則を適用する:

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

uu = u^{1 + 1}

= u^{1 + 1}

数を足す:

1 + 1 = 2

= u^{2}

= u - u^{2}

= u - u^{2}

= 1 - u + u - u^{2}

類似した元を足す:

- u + u = 0

= 1 - u^{2}

u^{2} = 1 - u^{2}

u^{2}

を左側に移動します

u^{2} = 1 - u^{2}

両辺に

u^{2}

を足す

u^{2} + u^{2} = 1 - u^{2} + u^{2}

簡素化

2 u^{2} = 1

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}

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

解を検算する

未定義の (特異) 点を求める:

u = 1

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

の分母をゼロに比較する

解く

1 - u = 0 : u = 1

1 - u = 0

1

を右側に移動します

1 - u = 0

両辺から

1

を引く

1 - u - 1 = 0 - 1

簡素化

- u = - 1

- u = - 1

以下で両辺を割る

- 1

- u = - 1

以下で両辺を割る

- 1

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

簡素化

u = 1

u = 1

以下の点は定義されていない

u = 1

未定義のポイントを解に組み合わせる:

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

代用を戻す

u = cos (x)

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.78539 \dots + 2 πn, x = 2 π - 0.78539 \dots + 2 πn, x = 2.35619 \dots + 2 πn, x = - 2.35619 \dots + 2 πn

グラフ

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