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人気のある三角関数 >

sec^2(θ)+2sec(θ)=8

解

sec^{2} (θ) + 2 sec (θ) = 8

解

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 1.82347 \dots + 2 πn, θ = - 1.82347 \dots + 2 πn

+1

度

θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 104.47751 \dots^{\circ} + 36 0^{\circ} n, θ = - 104.47751 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

sec^{2} (θ) + 2 sec (θ) = 8

置換で解く

sec^{2} (θ) + 2 sec (θ) = 8

仮定:

sec (θ) = u

u^{2} + 2 u = 8

u^{2} + 2 u = 8 : u = 2, u = - 4

u^{2} + 2 u = 8

8

を左側に移動します

u^{2} + 2 u = 8

両辺から

8

を引く

u^{2} + 2 u - 8 = 8 - 8

簡素化

u^{2} + 2 u - 8 = 0

u^{2} + 2 u - 8 = 0

解くとthe二次式

u^{2} + 2 u - 8 = 0

二次Equationの公式:

次の場合:

a = 1, b = 2, c = - 8

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

2^{2} - 4 \cdot 1 \cdot (- 8)

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 1 \cdot 8

数を乗じる:

4 \cdot 1 \cdot 8 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数を足す:

4 + 32 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

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

解を分離する

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

u = \frac{- 2 + 6}{2 \cdot 1} : 2

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

数を足す/引く:

- 2 + 6 = 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{- 2 - 6}{2 \cdot 1} : - 4

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

数を引く:

- 2 - 6 = - 8

= \frac{- 8}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 8}{2}

分数の規則を適用する:

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

= - \frac{8}{2}

数を割る:

\frac{8}{2} = 4

= - 4

二次equationの解:

u = 2, u = - 4

代用を戻す

u = sec (θ)

sec (θ) = 2, sec (θ) = - 4

sec (θ) = 2, sec (θ) = - 4

sec (θ) = 2 : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sec (θ) = 2

以下の一般解

sec (θ) = 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

sec (θ) = - 4 : θ = arcsec (- 4) + 2 πn, θ = - arcsec (- 4) + 2 πn

sec (θ) = - 4

三角関数の逆数プロパティを適用する

sec (θ) = - 4

以下の一般解

sec (θ) = - 4

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

θ = arcsec (- 4) + 2 πn, θ = - arcsec (- 4) + 2 πn

θ = arcsec (- 4) + 2 πn, θ = - arcsec (- 4) + 2 πn

すべての解を組み合わせる

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = arcsec (- 4) + 2 πn, θ = - arcsec (- 4) + 2 πn

10進法形式で解を証明する

θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn, θ = 1.82347 \dots + 2 πn, θ = - 1.82347 \dots + 2 πn

グラフ

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