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人気のある三角関数 >

5tan^2(x)-9sec(x)+2=0

解

5 tan^{2} (x) - 9 sec (x) + 2 = 0

解

x = 1.07121 \dots + 2 πn, x = 2 π - 1.07121 \dots + 2 πn

+1

度

x = 61.37616 \dots^{\circ} + 36 0^{\circ} n, x = 298.62383 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

5 tan^{2} (x) - 9 sec (x) + 2 = 0

三角関数の公式を使用して書き換える

2 + 5 tan^{2} (x) - 9 sec (x)

ピタゴラスの公式を使用する:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= 2 + 5 (sec^{2} (x) - 1) - 9 sec (x)

簡素化

2 + 5 (sec^{2} (x) - 1) - 9 sec (x) : 5 sec^{2} (x) - 9 sec (x) - 3

2 + 5 (sec^{2} (x) - 1) - 9 sec (x)

拡張

5 (sec^{2} (x) - 1) : 5 sec^{2} (x) - 5

5 (sec^{2} (x) - 1)

分配法則を適用する:

a (b - c) = ab - a c

a = 5, b = sec^{2} (x), c = 1

= 5 sec^{2} (x) - 5 \cdot 1

数を乗じる:

5 \cdot 1 = 5

= 5 sec^{2} (x) - 5

= 2 + 5 sec^{2} (x) - 5 - 9 sec (x)

簡素化

2 + 5 sec^{2} (x) - 5 - 9 sec (x) : 5 sec^{2} (x) - 9 sec (x) - 3

2 + 5 sec^{2} (x) - 5 - 9 sec (x)

条件のようなグループ

= 5 sec^{2} (x) - 9 sec (x) + 2 - 5

数を足す/引く:

2 - 5 = - 3

= 5 sec^{2} (x) - 9 sec (x) - 3

= 5 sec^{2} (x) - 9 sec (x) - 3

= 5 sec^{2} (x) - 9 sec (x) - 3

- 3 + 5 sec^{2} (x) - 9 sec (x) = 0

置換で解く

- 3 + 5 sec^{2} (x) - 9 sec (x) = 0

仮定:

sec (x) = u

- 3 + 5 u^{2} - 9 u = 0

- 3 + 5 u^{2} - 9 u = 0 : u = \frac{9 + 141}{10}, u = \frac{9 - 141}{10}

- 3 + 5 u^{2} - 9 u = 0

標準的な形式で書く

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

5 u^{2} - 9 u - 3 = 0

解くとthe二次式

5 u^{2} - 9 u - 3 = 0

二次Equationの公式:

次の場合:

a = 5, b = - 9, c = - 3

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

(- 9)^{2} - 4 \cdot 5 (- 3) = 141

(- 9)^{2} - 4 \cdot 5 (- 3)

規則を適用

- (- a) = a

= (- 9)^{2} + 4 \cdot 5 \cdot 3

指数の規則を適用する:

n

が偶数であれば

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

(- 9)^{2} = 9^{2}

= 9^{2} + 4 \cdot 5 \cdot 3

数を乗じる:

4 \cdot 5 \cdot 3 = 60

= 9^{2} + 60

9^{2} = 81

= 81 + 60

数を足す:

81 + 60 = 141

= 141

u_{1, 2} = \frac{- ( - 9 ) \pm 141}{2 \cdot 5}

解を分離する

u_{1} = \frac{- ( - 9 ) + 141}{2 \cdot 5}, u_{2} = \frac{- ( - 9 ) - 141}{2 \cdot 5}

u = \frac{- ( - 9 ) + 141}{2 \cdot 5} : \frac{9 + 141}{10}

\frac{- ( - 9 ) + 141}{2 \cdot 5}

規則を適用

- (- a) = a

= \frac{9 + 141}{2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{9 + 141}{10}

u = \frac{- ( - 9 ) - 141}{2 \cdot 5} : \frac{9 - 141}{10}

\frac{- ( - 9 ) - 141}{2 \cdot 5}

規則を適用

- (- a) = a

= \frac{9 - 141}{2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{9 - 141}{10}

二次equationの解:

u = \frac{9 + 141}{10}, u = \frac{9 - 141}{10}

代用を戻す

u = sec (x)

sec (x) = \frac{9 + 141}{10}, sec (x) = \frac{9 - 141}{10}

sec (x) = \frac{9 + 141}{10}, sec (x) = \frac{9 - 141}{10}

sec (x) = \frac{9 + 141}{10} : x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

sec (x) = \frac{9 + 141}{10}

三角関数の逆数プロパティを適用する

sec (x) = \frac{9 + 141}{10}

以下の一般解

sec (x) = \frac{9 + 141}{10}

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

sec (x) = \frac{9 - 141}{10} :

解なし

sec (x) = \frac{9 - 141}{10}

sec (x) \leq - 1 or sec (x) \geq 1

解なし

すべての解を組み合わせる

x = arcsec (\frac{9 + 141}{10}) + 2 πn, x = 2 π - arcsec (\frac{9 + 141}{10}) + 2 πn

10進法形式で解を証明する

x = 1.07121 \dots + 2 πn, x = 2 π - 1.07121 \dots + 2 πn

グラフ

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