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人気のある三角関数 >

3cos^2(x)-cos(x)=0

解

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

解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

+1

度

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

3 u^{2} - u = 0

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

3 u^{2} - u = 0

解くとthe二次式

3 u^{2} - u = 0

二次Equationの公式:

次の場合:

a = 3, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 3 \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 3 \cdot 0 = 0

4 \cdot 3 \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 3}

解を分離する

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

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

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

規則を適用

- (- a) = a

= \frac{1 + 1}{2 \cdot 3}

数を足す:

1 + 1 = 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}

u = \frac{- ( - 1 ) - 1}{2 \cdot 3} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{1 - 1}{2 \cdot 3}

数を引く:

1 - 1 = 0

= \frac{0}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{0}{6}

規則を適用

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

= 0

二次equationの解:

u = \frac{1}{3}, u = 0

代用を戻す

u = cos (x)

cos (x) = \frac{1}{3}, cos (x) = 0

cos (x) = \frac{1}{3}, cos (x) = 0

cos (x) = \frac{1}{3} : x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{3}

以下の一般解

cos (x) = \frac{1}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

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

すべての解を組み合わせる

x = arccos (\frac{1}{3}) + 2 πn, x = 2 π - arccos (\frac{1}{3}) + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

10進法形式で解を証明する

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

グラフ

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