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人気のある三角関数 >

cos^4(x)=8cos^4(x)-8cos^2(x)+1

解

cos^{4} (x) = 8 cos^{4} (x) - 8 cos^{2} (x) + 1

解

x = 2 πn, x = π + 2 πn, x = 1.18319 \dots + 2 πn, x = 2 π - 1.18319 \dots + 2 πn, x = 1.95839 \dots + 2 πn, x = - 1.95839 \dots + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 67.79234 \dots^{\circ} + 36 0^{\circ} n, x = 292.20765 \dots^{\circ} + 36 0^{\circ} n, x = 112.20765 \dots^{\circ} + 36 0^{\circ} n, x = - 112.20765 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

cos^{4} (x) = 8 cos^{4} (x) - 8 cos^{2} (x) + 1

置換で解く

cos^{4} (x) = 8 cos^{4} (x) - 8 cos^{2} (x) + 1

仮定:

cos (x) = u

u^{4} = 8 u^{4} - 8 u^{2} + 1

u^{4} = 8 u^{4} - 8 u^{2} + 1 : u = 1, u = - 1, u = \frac{1}{7}, u = - \frac{1}{7}

u^{4} = 8 u^{4} - 8 u^{2} + 1

辺を交換する

8 u^{4} - 8 u^{2} + 1 = u^{4}

u^{4}

を左側に移動します

8 u^{4} - 8 u^{2} + 1 = u^{4}

両辺から

u^{4}

を引く

8 u^{4} - 8 u^{2} + 1 - u^{4} = u^{4} - u^{4}

簡素化

7 u^{4} - 8 u^{2} + 1 = 0

7 u^{4} - 8 u^{2} + 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

7 v^{2} - 8 v + 1 = 0

解く

7 v^{2} - 8 v + 1 = 0 : v = 1, v = \frac{1}{7}

7 v^{2} - 8 v + 1 = 0

解くとthe二次式

7 v^{2} - 8 v + 1 = 0

二次Equationの公式:

次の場合:

a = 7, b = - 8, c = 1

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

v_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

(- 8)^{2} - 4 \cdot 7 \cdot 1 = 6

(- 8)^{2} - 4 \cdot 7 \cdot 1

指数の規則を適用する:

n

が偶数であれば

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

(- 8)^{2} = 8^{2}

= 8^{2} - 4 \cdot 7 \cdot 1

数を乗じる:

4 \cdot 7 \cdot 1 = 28

= 8^{2} - 28

8^{2} = 64

= 64 - 28

数を引く:

64 - 28 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

v_{1, 2} = \frac{- ( - 8 ) \pm 6}{2 \cdot 7}

解を分離する

v_{1} = \frac{- ( - 8 ) + 6}{2 \cdot 7}, v_{2} = \frac{- ( - 8 ) - 6}{2 \cdot 7}

v = \frac{- ( - 8 ) + 6}{2 \cdot 7} : 1

\frac{- ( - 8 ) + 6}{2 \cdot 7}

規則を適用

- (- a) = a

= \frac{8 + 6}{2 \cdot 7}

数を足す:

8 + 6 = 14

= \frac{14}{2 \cdot 7}

数を乗じる:

2 \cdot 7 = 14

= \frac{14}{14}

規則を適用

\frac{a}{a} = 1

= 1

v = \frac{- ( - 8 ) - 6}{2 \cdot 7} : \frac{1}{7}

\frac{- ( - 8 ) - 6}{2 \cdot 7}

規則を適用

- (- a) = a

= \frac{8 - 6}{2 \cdot 7}

数を引く:

8 - 6 = 2

= \frac{2}{2 \cdot 7}

数を乗じる:

2 \cdot 7 = 14

= \frac{2}{14}

共通因数を約分する:

2

= \frac{1}{7}

二次equationの解:

v = 1, v = \frac{1}{7}

v = 1, v = \frac{1}{7}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

解く

u^{2} = \frac{1}{7} : u = \frac{1}{7}, u = - \frac{1}{7}

u^{2} = \frac{1}{7}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = \frac{1}{7}, u = - \frac{1}{7}

解答は

u = 1, u = - 1, u = \frac{1}{7}, u = - \frac{1}{7}

代用を戻す

u = cos (x)

cos (x) = 1, cos (x) = - 1, cos (x) = \frac{1}{7}, cos (x) = - \frac{1}{7}

cos (x) = 1, cos (x) = - 1, cos (x) = \frac{1}{7}, cos (x) = - \frac{1}{7}

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) = - 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

cos (x) = \frac{1}{7} : x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (x) = \frac{1}{7}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{7}

以下の一般解

cos (x) = \frac{1}{7}

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

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn

cos (x) = - \frac{1}{7} : x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

cos (x) = - \frac{1}{7}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{1}{7}

以下の一般解

cos (x) = - \frac{1}{7}

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

x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn, x = arccos (\frac{1}{7}) + 2 πn, x = 2 π - arccos (\frac{1}{7}) + 2 πn, x = arccos (- \frac{1}{7}) + 2 πn, x = - arccos (- \frac{1}{7}) + 2 πn

10進法形式で解を証明する

x = 2 πn, x = π + 2 πn, x = 1.18319 \dots + 2 πn, x = 2 π - 1.18319 \dots + 2 πn, x = 1.95839 \dots + 2 πn, x = - 1.95839 \dots + 2 πn

グラフ

人気の例

sin^2(x+pi/2)= 3/4

sin^{2} (x + \frac{π}{2}) = \frac{3}{4}

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sin (2 x - \frac{π}{6}) = \frac{1}{2}

2cos(x)+1=3sec(x)

2 cos (x) + 1 = 3 sec (x)

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