ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

3sin^2(x)-5sin(x)+2=0

解

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

解

x = \frac{π}{2} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 41.81031 \dots^{\circ} + 36 0^{\circ} n, x = 138.18968 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

sin (x) = u

3 u^{2} - 5 u + 2 = 0

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

3 u^{2} - 5 u + 2 = 0

解くとthe二次式

3 u^{2} - 5 u + 2 = 0

二次Equationの公式:

次の場合:

a = 3, b = - 5, c = 2

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 3 \cdot 2}{2 \cdot 3}

(- 5)^{2} - 4 \cdot 3 \cdot 2 = 1

(- 5)^{2} - 4 \cdot 3 \cdot 2

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 3 \cdot 2

数を乗じる:

4 \cdot 3 \cdot 2 = 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 3}

解を分離する

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 3}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 3}

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

\frac{- ( - 5 ) + 1}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{5 + 1}{2 \cdot 3}

数を足す:

5 + 1 = 6

= \frac{6}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{6}{6}

規則を適用

\frac{a}{a} = 1

= 1

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

\frac{- ( - 5 ) - 1}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{5 - 1}{2 \cdot 3}

数を引く:

5 - 1 = 4

= \frac{4}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{4}{6}

共通因数を約分する:

2

= \frac{2}{3}

二次equationの解:

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

代用を戻す

u = sin (x)

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

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

以下の一般解

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = \frac{π}{2} + 2 πn, x = 0.72972 \dots + 2 πn, x = π - 0.72972 \dots + 2 πn

グラフ

人気の例

2cos(x)+sin(x)=0

2 cos (x) + sin (x) = 0

6sin^2(x)+12sin(x)+6=0

6 sin^{2} (x) + 12 sin (x) + 6 = 0

tan(1/2 x)+sqrt(3)=0

tan (\frac{1}{2} x) + 3 = 0

-arcsin(4x)= pi/4

- arcsin (4 x) = \frac{π}{4}

cos (x) = - 0.75