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人気のある三角関数 >

cos(2x)-sin(x)=1,0<= x<2pi

解

cos (2 x) - sin (x) = 1, 0 \leq x < 2 π

解

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0, x = π

+1

度

x = 21 0^{\circ}, x = 33 0^{\circ}, x = 0^{\circ}, x = 18 0^{\circ}

解答ステップ

cos (2 x) - sin (x) = 1, 0 \leq x < 2 π

両辺から

1

を引く

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 + cos (2 x) - sin (x)

2倍角の公式を使用:

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

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

簡素化

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

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

置換で解く

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

仮定:

sin (x) = u

- u - 2 u^{2} = 0

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

- u - 2 u^{2} = 0

標準的な形式で書く

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

- 2 u^{2} - u = 0

解くとthe二次式

- 2 u^{2} - u = 0

二次Equationの公式:

次の場合:

a = - 2, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(- 1)^{2} - 4 (- 2) \cdot 0 = 1

(- 1)^{2} - 4 (- 2) \cdot 0

規則を適用

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 0

(- 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 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

規則を適用

0 \cdot a = 0

= 0

= 1 + 0

数を足す:

1 + 0 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

括弧を削除する:

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

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

数を足す:

1 + 1 = 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{- 4}

分数の規則を適用する:

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

= - \frac{2}{4}

共通因数を約分する:

2

= - \frac{1}{2}

u = \frac{- ( - 1 ) - 1}{2 ( - 2 )} : 0

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

括弧を削除する:

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

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

数を引く:

1 - 1 = 0

= \frac{0}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{0}{- 4}

分数の規則を適用する:

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

= - \frac{0}{4}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

二次equationの解:

u = - \frac{1}{2}, u = 0

代用を戻す

u = sin (x)

sin (x) = - \frac{1}{2}, sin (x) = 0

sin (x) = - \frac{1}{2}, sin (x) = 0

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x < 2 π

以下の一般解

sin (x) = - \frac{1}{2}

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

範囲の解答

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

以下の一般解

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

範囲の解答

0 \leq x < 2 π

x = 0, x = π

すべての解を組み合わせる

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = 0, x = π

グラフ

人気の例

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sec^{2} (x) + 5 sec (x) - 14 = 0

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cos (θ) = - \frac{7}{8}, csc (θ)

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