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 >

12sin^2(x)-13sin(x)+3=0

解

12 sin^{2} (x) - 13 sin (x) + 3 = 0

解

x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn, x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn

+1

度

x = 48.59037 \dots^{\circ} + 36 0^{\circ} n, x = 131.40962 \dots^{\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

12 sin^{2} (x) - 13 sin (x) + 3 = 0

置換で解く

12 sin^{2} (x) - 13 sin (x) + 3 = 0

仮定:

sin (x) = u

12 u^{2} - 13 u + 3 = 0

12 u^{2} - 13 u + 3 = 0 : u = \frac{3}{4}, u = \frac{1}{3}

12 u^{2} - 13 u + 3 = 0

解くとthe二次式

12 u^{2} - 13 u + 3 = 0

二次Equationの公式:

次の場合:

a = 12, b = - 13, c = 3

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 12 \cdot 3}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 12 \cdot 3}{2 \cdot 12}

(- 13)^{2} - 4 \cdot 12 \cdot 3 = 5

(- 13)^{2} - 4 \cdot 12 \cdot 3

指数の規則を適用する:

n

が偶数であれば

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

(- 13)^{2} = 1 3^{2}

= 1 3^{2} - 4 \cdot 12 \cdot 3

数を乗じる:

4 \cdot 12 \cdot 3 = 144

= 1 3^{2} - 144

1 3^{2} = 169

= 169 - 144

数を引く:

169 - 144 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 13 ) \pm 5}{2 \cdot 12}

解を分離する

u_{1} = \frac{- ( - 13 ) + 5}{2 \cdot 12}, u_{2} = \frac{- ( - 13 ) - 5}{2 \cdot 12}

u = \frac{- ( - 13 ) + 5}{2 \cdot 12} : \frac{3}{4}

\frac{- ( - 13 ) + 5}{2 \cdot 12}

規則を適用

- (- a) = a

= \frac{13 + 5}{2 \cdot 12}

数を足す:

13 + 5 = 18

= \frac{18}{2 \cdot 12}

数を乗じる:

2 \cdot 12 = 24

= \frac{18}{24}

共通因数を約分する:

6

= \frac{3}{4}

u = \frac{- ( - 13 ) - 5}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 13 ) - 5}{2 \cdot 12}

規則を適用

- (- a) = a

= \frac{13 - 5}{2 \cdot 12}

数を引く:

13 - 5 = 8

= \frac{8}{2 \cdot 12}

数を乗じる:

2 \cdot 12 = 24

= \frac{8}{24}

共通因数を約分する:

8

= \frac{1}{3}

二次equationの解:

u = \frac{3}{4}, u = \frac{1}{3}

代用を戻す

u = sin (x)

sin (x) = \frac{3}{4}, sin (x) = \frac{1}{3}

sin (x) = \frac{3}{4}, sin (x) = \frac{1}{3}

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

sin (x) = \frac{3}{4}

三角関数の逆数プロパティを適用する

sin (x) = \frac{3}{4}

以下の一般解

sin (x) = \frac{3}{4}

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

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

x = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 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 = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 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 = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn, x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn

グラフ

人気の例

sin(θ)= 1/(sqrt(2))

sin (θ) = \frac{1}{2}

csc^{2} (x) - 1 = 0

csc(θ)=(2sqrt(3))/3

csc (θ) = \frac{2 3}{3}

- sin (2 x) = 0

30sin^2(x)=30+15cos(x)

30 sin^{2} (x) = 30 + 15 cos (x)