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人気のある三角関数 >

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

解

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

解

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = π + 2 πn

+1

度

x = 75.52248 \dots^{\circ} + 36 0^{\circ} n, x = 284.47751 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

3^{2} - 4 \cdot 4 (- 1) = 5

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

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 4 \cdot 1 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数を足す:

9 + 16 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 4}

解を分離する

u_{1} = \frac{- 3 + 5}{2 \cdot 4}, u_{2} = \frac{- 3 - 5}{2 \cdot 4}

u = \frac{- 3 + 5}{2 \cdot 4} : \frac{1}{4}

\frac{- 3 + 5}{2 \cdot 4}

数を足す/引く:

- 3 + 5 = 2

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

数を乗じる:

2 \cdot 4 = 8

= \frac{2}{8}

共通因数を約分する:

2

= \frac{1}{4}

u = \frac{- 3 - 5}{2 \cdot 4} : - 1

\frac{- 3 - 5}{2 \cdot 4}

数を引く:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 8}{8}

分数の規則を適用する:

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

= - \frac{8}{8}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

u = \frac{1}{4}, u = - 1

代用を戻す

u = cos (x)

cos (x) = \frac{1}{4}, cos (x) = - 1

cos (x) = \frac{1}{4}, cos (x) = - 1

cos (x) = \frac{1}{4} : x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (x) = \frac{1}{4}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{4}

以下の一般解

cos (x) = \frac{1}{4}

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

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn

cos (x) = - 1 : x = π + 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 = π + 2 πn

x = π + 2 πn

すべての解を組み合わせる

x = arccos (\frac{1}{4}) + 2 πn, x = 2 π - arccos (\frac{1}{4}) + 2 πn, x = π + 2 πn

10進法形式で解を証明する

x = 1.31811 \dots + 2 πn, x = 2 π - 1.31811 \dots + 2 πn, x = π + 2 πn

グラフ

人気の例

sin(4(x-pi/4))=0

sin (4 (x - \frac{π}{4})) = 0

tan(2θ)=1,0<= θ<= 360

tan (2 θ) = 1, 0^{\circ} \leq θ \leq 36 0^{\circ}

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