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人気のある三角関数 >

sin(x)(2sin(x)+1)=2

解

sin (x) (2 sin (x) + 1) = 2

解

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

+1

度

x = 51.33171 \dots^{\circ} + 36 0^{\circ} n, x = 128.66828 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

sin (x) (2 sin (x) + 1) = 2

置換で解く

sin (x) (2 sin (x) + 1) = 2

仮定:

sin (x) = u

u (2 u + 1) = 2

u (2 u + 1) = 2 : u = \frac{- 1 + 17}{4}, u = \frac{- 1 - 17}{4}

u (2 u + 1) = 2

拡張

u (2 u + 1) : 2 u^{2} + u

u (2 u + 1)

分配法則を適用する:

a (b + c) = ab + a c

a = u, b = 2 u, c = 1

= u \cdot 2 u + u \cdot 1

= 2 uu + 1 \cdot u

簡素化

2 uu + 1 \cdot u : 2 u^{2} + u

2 uu + 1 \cdot u

2 uu = 2 u^{2}

2 uu

指数の規則を適用する:

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

uu = u^{1 + 1}

= 2 u^{1 + 1}

数を足す:

1 + 1 = 2

= 2 u^{2}

1 \cdot u = u

1 \cdot u

乗算:

1 \cdot u = u

= u

= 2 u^{2} + u

= 2 u^{2} + u

2 u^{2} + u = 2

2

を左側に移動します

2 u^{2} + u = 2

両辺から

2

を引く

2 u^{2} + u - 2 = 2 - 2

簡素化

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 2, b = 1, c = - 2

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

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

1^{2} - 4 \cdot 2 (- 2) = 17

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

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 2)

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 2

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 1 + 16

数を足す:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 1 + 17}{2 \cdot 2}, u_{2} = \frac{- 1 - 17}{2 \cdot 2}

u = \frac{- 1 + 17}{2 \cdot 2} : \frac{- 1 + 17}{4}

\frac{- 1 + 17}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 1 + 17}{4}

u = \frac{- 1 - 17}{2 \cdot 2} : \frac{- 1 - 17}{4}

\frac{- 1 - 17}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 1 - 17}{4}

二次equationの解:

u = \frac{- 1 + 17}{4}, u = \frac{- 1 - 17}{4}

代用を戻す

u = sin (x)

sin (x) = \frac{- 1 + 17}{4}, sin (x) = \frac{- 1 - 17}{4}

sin (x) = \frac{- 1 + 17}{4}, sin (x) = \frac{- 1 - 17}{4}

sin (x) = \frac{- 1 + 17}{4} : x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{- 1 + 17}{4}

三角関数の逆数プロパティを適用する

sin (x) = \frac{- 1 + 17}{4}

以下の一般解

sin (x) = \frac{- 1 + 17}{4}

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

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{- 1 - 17}{4} :

解なし

sin (x) = \frac{- 1 - 17}{4}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

10進法形式で解を証明する

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

グラフ

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