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人気のある三角関数 >

2sin^2(x)-cos^2(x)-4sin(x)+2=0

解

2 sin^{2} (x) - cos^{2} (x) - 4 sin (x) + 2 = 0

解

x = \frac{π}{2} + 2 πn, x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn

+1

度

x = 9 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

2 sin^{2} (x) - cos^{2} (x) - 4 sin (x) + 2 = 0

三角関数の公式を使用して書き換える

2 - cos^{2} (x) + 2 sin^{2} (x) - 4 sin (x)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= 2 - (1 - sin^{2} (x)) + 2 sin^{2} (x) - 4 sin (x)

簡素化

2 - (1 - sin^{2} (x)) + 2 sin^{2} (x) - 4 sin (x) : 3 sin^{2} (x) - 4 sin (x) + 1

2 - (1 - sin^{2} (x)) + 2 sin^{2} (x) - 4 sin (x)

- (1 - sin^{2} (x)) : - 1 + sin^{2} (x)

- (1 - sin^{2} (x))

括弧を分配する

= - (1) - (- sin^{2} (x))

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

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

= 2 - 1 + sin^{2} (x) + 2 sin^{2} (x) - 4 sin (x)

簡素化

2 - 1 + sin^{2} (x) + 2 sin^{2} (x) - 4 sin (x) : 3 sin^{2} (x) - 4 sin (x) + 1

2 - 1 + sin^{2} (x) + 2 sin^{2} (x) - 4 sin (x)

類似した元を足す:

sin^{2} (x) + 2 sin^{2} (x) = 3 sin^{2} (x)

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

数を引く:

2 - 1 = 1

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

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

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

1 + 3 sin^{2} (x) - 4 sin (x) = 0

置換で解く

1 + 3 sin^{2} (x) - 4 sin (x) = 0

仮定:

sin (x) = u

1 + 3 u^{2} - 4 u = 0

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

1 + 3 u^{2} - 4 u = 0

標準的な形式で書く

a x^{2} + b x + c = 0

3 u^{2} - 4 u + 1 = 0

解くとthe二次式

3 u^{2} - 4 u + 1 = 0

二次Equationの公式:

次の場合:

a = 3, b = - 4, c = 1

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

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

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

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

指数の規則を適用する:

n

が偶数であれば

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

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

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

数を乗じる:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数を引く:

16 - 12 = 4

= 4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

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

解を分離する

u_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 3}, u_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 3}

u = \frac{- ( - 4 ) + 2}{2 \cdot 3} : 1

\frac{- ( - 4 ) + 2}{2 \cdot 3}

規則を適用

- (- a) = a

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

数を足す:

4 + 2 = 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{- ( - 4 ) - 2}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 4 ) - 2}{2 \cdot 3}

規則を適用

- (- a) = a

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

数を引く:

4 - 2 = 2

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

数を乗じる:

2 \cdot 3 = 6

= \frac{2}{6}

共通因数を約分する:

2

= \frac{1}{3}

二次equationの解:

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

代用を戻す

u = sin (x)

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

sin (x) = 1, sin (x) = \frac{1}{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{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}{2} + 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}{2} + 2 πn, x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn

グラフ

人気の例

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

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

1+cos^2(a)=3sin(x)cos(x)

1 + cos^{2} (a) = 3 sin (x) cos (x)

3cos(x)+sin(x)=cos(x)-2sin(x)

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tan^2(x)-sin(x)=tan^2(x)sin^2(x)

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