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人気のある三角関数 >

sec^2(θ)-9sec(θ)+20=0

解

sec^{2} (θ) - 9 sec (θ) + 20 = 0

解

θ = 1.36943 \dots + 2 πn, θ = 2 π - 1.36943 \dots + 2 πn, θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

+1

度

θ = 78.46304 \dots^{\circ} + 36 0^{\circ} n, θ = 281.53695 \dots^{\circ} + 36 0^{\circ} n, θ = 75.52248 \dots^{\circ} + 36 0^{\circ} n, θ = 284.47751 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

sec^{2} (θ) - 9 sec (θ) + 20 = 0

置換で解く

sec^{2} (θ) - 9 sec (θ) + 20 = 0

仮定:

sec (θ) = u

u^{2} - 9 u + 20 = 0

u^{2} - 9 u + 20 = 0 : u = 5, u = 4

u^{2} - 9 u + 20 = 0

解くとthe二次式

u^{2} - 9 u + 20 = 0

二次Equationの公式:

次の場合:

a = 1, b = - 9, c = 20

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

u_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 20}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot 20 = 1

(- 9)^{2} - 4 \cdot 1 \cdot 20

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 1 \cdot 20

数を乗じる:

4 \cdot 1 \cdot 20 = 80

= 9^{2} - 80

9^{2} = 81

= 81 - 80

数を引く:

81 - 80 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

u = \frac{- ( - 9 ) + 1}{2 \cdot 1} : 5

\frac{- ( - 9 ) + 1}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{9 + 1}{2 \cdot 1}

数を足す:

9 + 1 = 10

= \frac{10}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{10}{2}

数を割る:

\frac{10}{2} = 5

= 5

u = \frac{- ( - 9 ) - 1}{2 \cdot 1} : 4

\frac{- ( - 9 ) - 1}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{9 - 1}{2 \cdot 1}

数を引く:

9 - 1 = 8

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

数を乗じる:

2 \cdot 1 = 2

= \frac{8}{2}

数を割る:

\frac{8}{2} = 4

= 4

二次equationの解:

u = 5, u = 4

代用を戻す

u = sec (θ)

sec (θ) = 5, sec (θ) = 4

sec (θ) = 5, sec (θ) = 4

sec (θ) = 5 : θ = arcsec (5) + 2 πn, θ = 2 π - arcsec (5) + 2 πn

sec (θ) = 5

三角関数の逆数プロパティを適用する

sec (θ) = 5

以下の一般解

sec (θ) = 5

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

θ = arcsec (5) + 2 πn, θ = 2 π - arcsec (5) + 2 πn

θ = arcsec (5) + 2 πn, θ = 2 π - arcsec (5) + 2 πn

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

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

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

すべての解を組み合わせる

θ = arcsec (5) + 2 πn, θ = 2 π - arcsec (5) + 2 πn, θ = arcsec (4) + 2 πn, θ = 2 π - arcsec (4) + 2 πn

10進法形式で解を証明する

θ = 1.36943 \dots + 2 πn, θ = 2 π - 1.36943 \dots + 2 πn, θ = 1.31811 \dots + 2 πn, θ = 2 π - 1.31811 \dots + 2 πn

グラフ

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