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人気のある三角関数 >

3cos^2(x)-4cos(x)+1=0

解

3 cos^{2} (x) - 4 cos (x) + 1 = 0

解

x = 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

+1

度

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

解答ステップ

3 cos^{2} (x) - 4 cos (x) + 1 = 0

置換で解く

3 cos^{2} (x) - 4 cos (x) + 1 = 0

仮定:

cos (x) = u

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

3 u^{2} - 4 u + 1 = 0 : u = 1, u = \frac{1}{3}

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 3, b = - 4, c = 1

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

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

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

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

指数の規則を適用する:

n

が偶数であれば

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

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

= 4^{2} - 4 \cdot 3 \cdot 1

数を乗じる:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数を引く:

16 - 12 = 4

= 4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

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

解を分離する

u_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 3}, u_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 3}

u = \frac{- ( - 4 ) + 2}{2 \cdot 3} : 1

\frac{- ( - 4 ) + 2}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{4 + 2}{2 \cdot 3}

数を足す:

4 + 2 = 6

= \frac{6}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{6}{6}

規則を適用

\frac{a}{a} = 1

= 1

u = \frac{- ( - 4 ) - 2}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 4 ) - 2}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{4 - 2}{2 \cdot 3}

数を引く:

4 - 2 = 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}

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = 1 : x = 2 πn

cos (x) = 1

以下の一般解

cos (x) = 1

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 = 0 + 2 πn

x = 0 + 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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 2 πn, x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn

グラフ

人気の例

cos (x) = 0.1

3 tan (x) = 2.4568

cos (x) = 1.5

2/(sqrt(3))cos(3θ)=1,0<= θ<= 2pi

\frac{2}{3} cos (3 θ) = 1, 0 \leq θ \leq 2 π

8cos(B)+sqrt(52)=0

8 cos (B) + 52 = 0