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人気のある三角関数 >

2csc^2(x)+5cot(x)=5

解

2 csc^{2} (x) + 5 cot (x) = 5

解

x = 1.10714 \dots + πn, x = 2.81984 \dots + πn

+1

度

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 161.56505 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

2 csc^{2} (x) + 5 cot (x) = 5

両辺から

5

を引く

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

三角関数の公式を使用して書き換える

- 5 + 2 csc^{2} (x) + 5 cot (x)

ピタゴラスの公式を使用する:

csc^{2} (x) = 1 + cot^{2} (x)

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

簡素化

- 5 + 2 (1 + cot^{2} (x)) + 5 cot (x) : 2 cot^{2} (x) + 5 cot (x) - 3

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

拡張

2 (1 + cot^{2} (x)) : 2 + 2 cot^{2} (x)

2 (1 + cot^{2} (x))

分配法則を適用する:

a (b + c) = ab + a c

a = 2, b = 1, c = cot^{2} (x)

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

数を乗じる:

2 \cdot 1 = 2

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

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

数を足す/引く:

- 5 + 2 = - 3

= 2 cot^{2} (x) + 5 cot (x) - 3

= 2 cot^{2} (x) + 5 cot (x) - 3

- 3 + 2 cot^{2} (x) + 5 cot (x) = 0

置換で解く

- 3 + 2 cot^{2} (x) + 5 cot (x) = 0

仮定:

cot (x) = u

- 3 + 2 u^{2} + 5 u = 0

- 3 + 2 u^{2} + 5 u = 0 : u = \frac{1}{2}, u = - 3

- 3 + 2 u^{2} + 5 u = 0

標準的な形式で書く

a x^{2} + b x + c = 0

2 u^{2} + 5 u - 3 = 0

解くとthe二次式

2 u^{2} + 5 u - 3 = 0

二次Equationの公式:

次の場合:

a = 2, b = 5, c = - 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 3 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 3) = 7

5^{2} - 4 \cdot 2 (- 3)

規則を適用

- (- a) = a

= 5^{2} + 4 \cdot 2 \cdot 3

数を乗じる:

4 \cdot 2 \cdot 3 = 24

= 5^{2} + 24

5^{2} = 25

= 25 + 24

数を足す:

25 + 24 = 49

= 49

数を因数に分解する:

49 = 7^{2}

= 7^{2}

累乗根の規則を適用する:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 5 \pm 7}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 5 + 7}{2 \cdot 2}, u_{2} = \frac{- 5 - 7}{2 \cdot 2}

u = \frac{- 5 + 7}{2 \cdot 2} : \frac{1}{2}

\frac{- 5 + 7}{2 \cdot 2}

数を足す/引く:

- 5 + 7 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 5 - 7}{2 \cdot 2} : - 3

\frac{- 5 - 7}{2 \cdot 2}

数を引く:

- 5 - 7 = - 12

= \frac{- 12}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 12}{4}

分数の規則を適用する:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{12}{4}

数を割る:

\frac{12}{4} = 3

= - 3

二次equationの解:

u = \frac{1}{2}, u = - 3

代用を戻す

u = cot (x)

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

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

cot (x) = \frac{1}{2} : x = arccot (\frac{1}{2}) + πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

x = arccot (\frac{1}{2}) + πn

x = arccot (\frac{1}{2}) + πn

cot (x) = - 3 : x = arccot (- 3) + πn

cot (x) = - 3

三角関数の逆数プロパティを適用する

cot (x) = - 3

以下の一般解

cot (x) = - 3

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

x = arccot (- 3) + πn

x = arccot (- 3) + πn

すべての解を組み合わせる

x = arccot (\frac{1}{2}) + πn, x = arccot (- 3) + πn

10進法形式で解を証明する

x = 1.10714 \dots + πn, x = 2.81984 \dots + πn

グラフ

人気の例

(1+tan(x))(5sin(x)-2)=0

(1 + tan (x)) (5 sin (x) - 2) = 0

arccos(e-x)=pi^3

arccos (e - x) = π^{3}

sin (a) = \frac{20}{25}

1+cos(x)=sqrt(3)-sin(x)

1 + cos (x) = 3 - sin (x)

3 = 7 cos (\frac{π}{3} t)