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人気のある三角関数 >

tan^2(x)+cot^2(x)-2=0

解

tan^{2} (x) + cot^{2} (x) - 2 = 0

解

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

+1

度

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

解答ステップ

tan^{2} (x) + cot^{2} (x) - 2 = 0

三角関数の公式を使用して書き換える

- 2 + cot^{2} (x) + tan^{2} (x)

基本的な三角関数の公式を使用する:

tan (x) = \frac{1}{cot ( x )}

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

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

(\frac{1}{cot ( x )})^{2}

指数の規則を適用する:

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

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

規則を適用

1^{a} = 1

1^{2} = 1

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

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

- 2 + cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

置換で解く

- 2 + cot^{2} (x) + \frac{1}{cot ^{2} ( x )} = 0

仮定:

cot (x) = u

- 2 + u^{2} + \frac{1}{u ^{2}} = 0

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

- 2 + u^{2} + \frac{1}{u ^{2}} = 0

以下で両辺を乗じる:

u^{2}

- 2 + u^{2} + \frac{1}{u ^{2}} = 0

以下で両辺を乗じる:

u^{2}

- 2 u^{2} + u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

簡素化

- 2 u^{2} + u^{2} u^{2} + \frac{1}{u ^{2}} u^{2} = 0 \cdot u^{2}

簡素化

u^{2} u^{2} : u^{4}

u^{2} u^{2}

指数の規則を適用する:

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

u^{2} u^{2} = u^{2 + 2}

= u^{2 + 2}

数を足す:

2 + 2 = 4

= u^{4}

簡素化

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

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

分数を乗じる:

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

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

共通因数を約分する:

u^{2}

= 1

簡素化

0 \cdot u^{2} : 0

0 \cdot u^{2}

規則を適用

0 \cdot a = 0

= 0

- 2 u^{2} + u^{4} + 1 = 0

- 2 u^{2} + u^{4} + 1 = 0

- 2 u^{2} + u^{4} + 1 = 0

解く

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

- 2 u^{2} + u^{4} + 1 = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{4} - 2 u^{2} + 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

v^{2} - 2 v + 1 = 0

解く

v^{2} - 2 v + 1 = 0 : v = 1

v^{2} - 2 v + 1 = 0

解くとthe二次式

v^{2} - 2 v + 1 = 0

二次Equationの公式:

次の場合:

a = 1, b = - 2, c = 1

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

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

指数の規則を適用する:

n

が偶数であれば

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

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

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

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

数を引く:

4 - 4 = 0

= 0

v_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

v = \frac{- ( - 2 )}{2 \cdot 1}

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

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

規則を適用

- (- a) = a

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

数を乗じる:

2 \cdot 1 = 2

= \frac{2}{2}

規則を適用

\frac{a}{a} = 1

= 1

v = 1

二次equationの解:

v = 1

v = 1

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

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

u^{2} = 1

x^{2} = f (a)

の場合, 解は

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

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

解答は

u = 1, u = - 1

u = 1, u = - 1

解を検算する

未定義の (特異) 点を求める:

u = 0

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

の分母をゼロに比較する

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

以下の点は定義されていない

u = 0

未定義のポイントを解に組み合わせる:

u = 1, u = - 1

代用を戻す

u = cot (x)

cot (x) = 1, cot (x) = - 1

cot (x) = 1, cot (x) = - 1

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

以下の一般解

cot (x) = 1

cot (x)

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

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

cot (x) = - 1

以下の一般解

cot (x) = - 1

cot (x)

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

すべての解を組み合わせる

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

グラフ

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