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人気のある三角関数 >

8sin^3(x)-4sin^2(x)-6sin(x)+3=0

解

8 sin^{3} (x) - 4 sin^{2} (x) - 6 sin (x) + 3 = 0

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n

解答ステップ

8 sin^{3} (x) - 4 sin^{2} (x) - 6 sin (x) + 3 = 0

置換で解く

8 sin^{3} (x) - 4 sin^{2} (x) - 6 sin (x) + 3 = 0

仮定:

sin (x) = u

8 u^{3} - 4 u^{2} - 6 u + 3 = 0

8 u^{3} - 4 u^{2} - 6 u + 3 = 0 : u = \frac{1}{2}, u = - \frac{3}{2}, u = \frac{3}{2}

8 u^{3} - 4 u^{2} - 6 u + 3 = 0

因数

8 u^{3} - 4 u^{2} - 6 u + 3 : (2 u - 1) (2 u + 3) (2 u - 3)

8 u^{3} - 4 u^{2} - 6 u + 3

= (8 u^{3} - 4 u^{2}) + (- 6 u + 3)

- 3

を

- 6 u + 3 : - 3 (2 u - 1)

からくくり出す

- 6 u + 3

6

を書き換え

3 \cdot 2

= - 3 \cdot 2 u + 3

共通項をくくり出す

- 3

= - 3 (2 u - 1)

4 u^{2}

を

8 u^{3} - 4 u^{2} : 4 u^{2} (2 u - 1)

からくくり出す

8 u^{3} - 4 u^{2}

指数の規則を適用する:

a^{b + c} = a^{b} a^{c}

u^{3} = u u^{2}

= 8 u u^{2} - 4 u^{2}

8

を書き換え

4 \cdot 2

= 4 \cdot 2 u u^{2} - 4 u^{2}

共通項をくくり出す

4 u^{2}

= 4 u^{2} (2 u - 1)

= - 3 (2 u - 1) + 4 u^{2} (2 u - 1)

共通項をくくり出す

2 u - 1

= (2 u - 1) (4 u^{2} - 3)

因数

4 u^{2} - 3 : (2 u + 3) (2 u - 3)

4 u^{2} - 3

4 u^{2} - 3

を書き換え

(2 u)^{2} - (3)^{2}

4 u^{2} - 3

4

を書き換え

2^{2}

= 2^{2} u^{2} - 3

累乗根の規則を適用する:

a = (a)^{2}

3 = (3)^{2}

= 2^{2} u^{2} - (3)^{2}

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

2^{2} u^{2} = (2 u)^{2}

= (2 u)^{2} - (3)^{2}

= (2 u)^{2} - (3)^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

(2 u)^{2} - (3)^{2} = (2 u + 3) (2 u - 3)

= (2 u + 3) (2 u - 3)

= (2 u - 1) (2 u + 3) (2 u - 3)

(2 u - 1) (2 u + 3) (2 u - 3) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

2 u - 1 = 0 or 2 u + 3 = 0 or 2 u - 3 = 0

解く

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

1

を右側に移動します

2 u - 1 = 0

両辺に

1

を足す

2 u - 1 + 1 = 0 + 1

簡素化

2 u = 1

2 u = 1

以下で両辺を割る

2

2 u = 1

以下で両辺を割る

2

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

簡素化

u = \frac{1}{2}

u = \frac{1}{2}

解く

2 u + 3 = 0 : u = - \frac{3}{2}

2 u + 3 = 0

3

を右側に移動します

2 u + 3 = 0

両辺から

3

を引く

2 u + 3 - 3 = 0 - 3

簡素化

2 u = - 3

2 u = - 3

以下で両辺を割る

2

2 u = - 3

以下で両辺を割る

2

\frac{2 u}{2} = \frac{- 3}{2}

簡素化

u = - \frac{3}{2}

u = - \frac{3}{2}

解く

2 u - 3 = 0 : u = \frac{3}{2}

2 u - 3 = 0

3

を右側に移動します

2 u - 3 = 0

両辺に

3

を足す

2 u - 3 + 3 = 0 + 3

簡素化

2 u = 3

2 u = 3

以下で両辺を割る

2

2 u = 3

以下で両辺を割る

2

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

簡素化

u = \frac{3}{2}

u = \frac{3}{2}

解答は

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

代用を戻す

u = sin (x)

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

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

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{3}{2} : x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sin (x) = - \frac{3}{2}

以下の一般解

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

x = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

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

以下の一般解

sin (x) = \frac{3}{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}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{2 π}{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 = \frac{4 π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn

グラフ

人気の例

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tan(2θ)+tan(2θ)=tan(4θ)

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