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人気のある三角関数 >

tan^2(x)=sec(x)

解

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

解

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

+1

度

x = 51.82729 \dots^{\circ} + 36 0^{\circ} n, x = 308.17270 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

sec (x)

を引く

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

三角関数の公式を使用して書き換える

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

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

- 1 - sec (x) + sec^{2} (x) = 0

置換で解く

- 1 - sec (x) + sec^{2} (x) = 0

仮定:

sec (x) = u

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

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

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

標準的な形式で書く

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

u^{2} - u - 1 = 0

解くとthe二次式

u^{2} - u - 1 = 0

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

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

(- 1)^{2} = 1^{2}

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

数を足す:

1 + 4 = 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} : \frac{1 + 5}{2}

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

規則を適用

- (- a) = a

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

数を乗じる:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

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

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

規則を適用

- (- a) = a

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

数を乗じる:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

二次equationの解:

u = \frac{1 + 5}{2}, u = \frac{1 - 5}{2}

代用を戻す

u = sec (x)

sec (x) = \frac{1 + 5}{2}, sec (x) = \frac{1 - 5}{2}

sec (x) = \frac{1 + 5}{2}, sec (x) = \frac{1 - 5}{2}

sec (x) = \frac{1 + 5}{2} : x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

sec (x) = \frac{1 + 5}{2}

三角関数の逆数プロパティを適用する

sec (x) = \frac{1 + 5}{2}

以下の一般解

sec (x) = \frac{1 + 5}{2}

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

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

x = arcsec (\frac{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

sec (x) = \frac{1 - 5}{2} :

解なし

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

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{1 + 5}{2}) + 2 πn, x = 2 π - arcsec (\frac{1 + 5}{2}) + 2 πn

10進法形式で解を証明する

x = 0.90455 \dots + 2 πn, x = 2 π - 0.90455 \dots + 2 πn

グラフ

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