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人気のある三角関数 >

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

解

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

解

x = \frac{π}{3}, x = \frac{5 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

+1

度

x = 6 0^{\circ}, x = 30 0^{\circ}, x = 9 0^{\circ}, x = 27 0^{\circ}

解答ステップ

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

両辺から

- 1

を引く

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

三角関数の公式を使用して書き換える

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

2倍角の公式を使用:

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

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

簡素化

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

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

条件のようなグループ

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

1 - 1 = 0

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

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

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

置換で解く

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

仮定:

cos (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 \cdot 2 \cdot 0}{2 \cdot 2}

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

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

(- 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

数を引く:

1 - 0 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

規則を適用

- (- 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}

共通因数を約分する:

2

= \frac{1}{2}

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

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

規則を適用

- (- 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}

規則を適用

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

= 0

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{1}{2}, 0 \leq x < 2 π

以下の一般解

cos (x) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

範囲の解答

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

以下の一般解

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

範囲の解答

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

すべての解を組み合わせる

x = \frac{π}{3}, x = \frac{5 π}{3}, x = \frac{π}{2}, x = \frac{3 π}{2}

グラフ

人気の例

sin(x)=-1/(sqrt(10))

sin (x) = - \frac{1}{10}

sec^2(θ)+sec(θ)=2,2pi<= θ<= 3pi

sec^{2} (θ) + sec (θ) = 2, 2 π \leq θ \leq 3 π

6 sec^{2} (x) = 0

sqrt(3-2cos(x))=2

3 - 2 cos (x) = 2

tan (θ) = - \frac{8}{5}