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人気のある三角関数 >

6sin^2(x)-5sin(x)+1=0

解

6 sin^{2} (x) - 5 sin (x) + 1 = 0

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 = 160.52877 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

6 sin^{2} (x) - 5 sin (x) + 1 = 0

置換で解く

6 sin^{2} (x) - 5 sin (x) + 1 = 0

仮定:

sin (x) = u

6 u^{2} - 5 u + 1 = 0

6 u^{2} - 5 u + 1 = 0 : u = \frac{1}{2}, u = \frac{1}{3}

6 u^{2} - 5 u + 1 = 0

解くとthe二次式

6 u^{2} - 5 u + 1 = 0

二次Equationの公式:

次の場合:

a = 6, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

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

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

指数の規則を適用する:

n

が偶数であれば

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

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

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

数を乗じる:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数を引く:

25 - 24 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 6}

解を分離する

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 6}

u = \frac{- ( - 5 ) + 1}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 6}

規則を適用

- (- a) = a

= \frac{5 + 1}{2 \cdot 6}

数を足す:

5 + 1 = 6

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

数を乗じる:

2 \cdot 6 = 12

= \frac{6}{12}

共通因数を約分する:

6

= \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 6} : \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 6}

規則を適用

- (- a) = a

= \frac{5 - 1}{2 \cdot 6}

数を引く:

5 - 1 = 4

= \frac{4}{2 \cdot 6}

数を乗じる:

2 \cdot 6 = 12

= \frac{4}{12}

共通因数を約分する:

4

= \frac{1}{3}

二次equationの解:

u = \frac{1}{2}, u = \frac{1}{3}

代用を戻す

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = \frac{1}{3}

sin (x) = \frac{1}{2}, sin (x) = \frac{1}{3}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

以下の一般解

sin (x) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{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

sin (x) = \frac{1}{3} : x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

sin (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1}{3}

以下の一般解

sin (x) = \frac{1}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{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 = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{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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