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人気のある三角関数 >

cot^2(x)-6cot(x)+5=0,0<= x<= 2pi

解

cot^{2} (x) - 6 cot (x) + 5 = 0, 0 \leq x \leq 2 π

解

x = 0.19739 \dots, x = 0.19739 \dots + π, x = \frac{π}{4}, x = \frac{5 π}{4}

+1

度

x = 11.30993 \dots^{\circ}, x = 191.30993 \dots^{\circ}, x = 4 5^{\circ}, x = 22 5^{\circ}

解答ステップ

cot^{2} (x) - 6 cot (x) + 5 = 0, 0 \leq x \leq 2 π

置換で解く

cot^{2} (x) - 6 cot (x) + 5 = 0

仮定:

cot (x) = u

u^{2} - 6 u + 5 = 0

u^{2} - 6 u + 5 = 0 : u = 5, u = 1

u^{2} - 6 u + 5 = 0

解くとthe二次式

u^{2} - 6 u + 5 = 0

二次Equationの公式:

次の場合:

a = 1, b = - 6, c = 5

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

(- 6)^{2} - 4 \cdot 1 \cdot 5

指数の規則を適用する:

n

が偶数であれば

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

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 1 \cdot 5

数を乗じる:

4 \cdot 1 \cdot 5 = 20

= 6^{2} - 20

6^{2} = 36

= 36 - 20

数を引く:

36 - 20 = 16

= 16

数を因数に分解する:

16 = 4^{2}

= 4^{2}

累乗根の規則を適用する:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 6 ) \pm 4}{2 \cdot 1}

解を分離する

u_{1} = \frac{- ( - 6 ) + 4}{2 \cdot 1}, u_{2} = \frac{- ( - 6 ) - 4}{2 \cdot 1}

u = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

\frac{- ( - 6 ) + 4}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{6 + 4}{2 \cdot 1}

数を足す:

6 + 4 = 10

= \frac{10}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{10}{2}

数を割る:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

\frac{- ( - 6 ) - 4}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{6 - 4}{2 \cdot 1}

数を引く:

6 - 4 = 2

= \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

二次equationの解:

u = 5, u = 1

代用を戻す

u = cot (x)

cot (x) = 5, cot (x) = 1

cot (x) = 5, cot (x) = 1

cot (x) = 5, 0 \leq x \leq 2 π : x = arccot (5), x = arccot (5) + π

cot (x) = 5, 0 \leq x \leq 2 π

三角関数の逆数プロパティを適用する

cot (x) = 5

以下の一般解

cot (x) = 5

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (5) + πn

x = arccot (5) + πn

範囲の解答

0 \leq x \leq 2 π

x = arccot (5), x = arccot (5) + π

cot (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

cot (x) = 1, 0 \leq x \leq 2 π

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

範囲の解答

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{5 π}{4}

すべての解を組み合わせる

x = arccot (5), x = arccot (5) + π, x = \frac{π}{4}, x = \frac{5 π}{4}

10進法形式で解を証明する

x = 0.19739 \dots, x = 0.19739 \dots + π, x = \frac{π}{4}, x = \frac{5 π}{4}

グラフ

人気の例

9tan(θ)+13=2tan(θ)+6

9 tan (θ) + 13 = 2 tan (θ) + 6

2 = 3 sin (∣ 5 x ∣) + 1

sin (θ) = - \frac{1}{3}

sin (θ) = - \frac{1}{4}

sin(θ)cos(θ)= 1/2

sin (θ) cos (θ) = \frac{1}{2}