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人気のある三角関数 >

2cot^2(x)-csc^2(x)+csc(x)=4

解

2 cot^{2} (x) - csc^{2} (x) + csc (x) = 4

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

2 cot^{2} (x) - csc^{2} (x) + csc (x) = 4

両辺から

4

を引く

2 cot^{2} (x) - csc^{2} (x) + csc (x) - 4 = 0

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

- 4 + csc (x) - csc^{2} (x) + 2 (csc^{2} (x) - 1) : csc^{2} (x) + csc (x) - 6

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

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

簡素化

- 4 + csc (x) - csc^{2} (x) + 2 csc^{2} (x) - 2 : csc^{2} (x) + csc (x) - 6

- 4 + csc (x) - csc^{2} (x) + 2 csc^{2} (x) - 2

類似した元を足す:

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

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

条件のようなグループ

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

数を引く:

- 4 - 2 = - 6

= csc^{2} (x) + csc (x) - 6

= csc^{2} (x) + csc (x) - 6

= csc^{2} (x) + csc (x) - 6

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

置換で解く

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

仮定:

csc (x) = u

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

- 6 + u + u^{2} = 0 : u = 2, u = - 3

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

標準的な形式で書く

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

u^{2} + u - 6 = 0

解くとthe二次式

u^{2} + u - 6 = 0

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 6)

規則を適用

- (- a) = a

= 1 + 4 \cdot 1 \cdot 6

数を乗じる:

4 \cdot 1 \cdot 6 = 24

= 1 + 24

数を足す:

1 + 24 = 25

= 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

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

解を分離する

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

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

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

数を足す/引く:

- 1 + 5 = 4

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

数を乗じる:

2 \cdot 1 = 2

= \frac{4}{2}

数を割る:

\frac{4}{2} = 2

= 2

u = \frac{- 1 - 5}{2 \cdot 1} : - 3

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

数を引く:

- 1 - 5 = - 6

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

数を乗じる:

2 \cdot 1 = 2

= \frac{- 6}{2}

分数の規則を適用する:

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

= - \frac{6}{2}

数を割る:

\frac{6}{2} = 3

= - 3

二次equationの解:

u = 2, u = - 3

代用を戻す

u = csc (x)

csc (x) = 2, csc (x) = - 3

csc (x) = 2, csc (x) = - 3

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

以下の一般解

csc (x) = 2

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = - 3 : x = arccsc (- 3) + 2 πn, x = π + arccsc (3) + 2 πn

csc (x) = - 3

三角関数の逆数プロパティを適用する

csc (x) = - 3

以下の一般解

csc (x) = - 3

csc (x) = - a \Rightarrow x = arccsc (- a) + 2 πn, x = π + arccsc (a) + 2 πn

x = arccsc (- 3) + 2 πn, x = π + arccsc (3) + 2 πn

x = arccsc (- 3) + 2 πn, x = π + arccsc (3) + 2 πn

すべての解を組み合わせる

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = arccsc (- 3) + 2 πn, x = π + arccsc (3) + 2 πn

10進法形式で解を証明する

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

グラフ

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