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人気のある三角関数 >

2cos(2x)-sin(x)-1=0

解

2 cos (2 x) - sin (x) - 1 = 0

解

x = - 0.69500 \dots + 2 πn, x = π + 0.69500 \dots + 2 πn, x = 0.40105 \dots + 2 πn, x = π - 0.40105 \dots + 2 πn

+1

度

x = - 39.82077 \dots^{\circ} + 36 0^{\circ} n, x = 219.82077 \dots^{\circ} + 36 0^{\circ} n, x = 22.97865 \dots^{\circ} + 36 0^{\circ} n, x = 157.02134 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

2 cos (2 x) - sin (x) - 1 = 0

三角関数の公式を使用して書き換える

- 1 - sin (x) + 2 cos (2 x)

2倍角の公式を使用:

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

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = 2 sin^{2} (x)

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

簡素化

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) : 2 - 4 sin^{2} (x)

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

数を乗じる:

2 \cdot 1 = 2

= 2 - 2 \cdot 2 sin^{2} (x)

数を乗じる:

2 \cdot 2 = 4

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

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

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

簡素化

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

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

条件のようなグループ

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

数を足す/引く:

- 1 + 2 = 1

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

1 - u - 4 u^{2} = 0 : u = - \frac{1 + 17}{8}, u = \frac{17 - 1}{8}

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

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

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

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 4 \cdot 1 = 16

4 \cdot 4 \cdot 1

数を乗じる:

4 \cdot 4 \cdot 1 = 16

= 16

= 1 + 16

数を足す:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 ( - 4 )}

解を分離する

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

u = \frac{- ( - 1 ) + 17}{2 ( - 4 )} : - \frac{1 + 17}{8}

\frac{- ( - 1 ) + 17}{2 ( - 4 )}

括弧を削除する:

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

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

数を乗じる:

2 \cdot 4 = 8

= \frac{1 + 17}{- 8}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1 + 17}{8}

u = \frac{- ( - 1 ) - 17}{2 ( - 4 )} : \frac{17 - 1}{8}

\frac{- ( - 1 ) - 17}{2 ( - 4 )}

括弧を削除する:

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

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

数を乗じる:

2 \cdot 4 = 8

= \frac{1 - 17}{- 8}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

1 - 17 = - (17 - 1)

= \frac{17 - 1}{8}

二次equationの解:

u = - \frac{1 + 17}{8}, u = \frac{17 - 1}{8}

代用を戻す

u = sin (x)

sin (x) = - \frac{1 + 17}{8}, sin (x) = \frac{17 - 1}{8}

sin (x) = - \frac{1 + 17}{8}, sin (x) = \frac{17 - 1}{8}

sin (x) = - \frac{1 + 17}{8} : x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn

sin (x) = - \frac{1 + 17}{8}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{1 + 17}{8}

以下の一般解

sin (x) = - \frac{1 + 17}{8}

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

x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn

x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn

sin (x) = \frac{17 - 1}{8} : x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

sin (x) = \frac{17 - 1}{8}

三角関数の逆数プロパティを適用する

sin (x) = \frac{17 - 1}{8}

以下の一般解

sin (x) = \frac{17 - 1}{8}

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

x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

すべての解を組み合わせる

x = arcsin (- \frac{1 + 17}{8}) + 2 πn, x = π + arcsin (\frac{1 + 17}{8}) + 2 πn, x = arcsin (\frac{17 - 1}{8}) + 2 πn, x = π - arcsin (\frac{17 - 1}{8}) + 2 πn

10進法形式で解を証明する

x = - 0.69500 \dots + 2 πn, x = π + 0.69500 \dots + 2 πn, x = 0.40105 \dots + 2 πn, x = π - 0.40105 \dots + 2 πn

グラフ

人気の例

2cos^2(θ)-2sin^2(θ)=sqrt(2)

2 cos^{2} (θ) - 2 sin^{2} (θ) = 2

cot(θ)+3csc(θ)=5

cot (θ) + 3 csc (θ) = 5

2cos^2(θ)-2sin^2(θ)=sqrt(3)

2 cos^{2} (θ) - 2 sin^{2} (θ) = 3

1 - ∣ cos (x) ∣ = 0

sin (θ) = - 0.789