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人気のある >

cos^4(x)=1-sin^4(x)

解

cos^{4} (x) = 1 - sin^{4} (x)

解

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

+1

度

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

cos^{4} (x) = 1 - sin^{4} (x)

両辺から

1 - sin^{4} (x)

を引く

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

指数の規則を適用する:

a^{b} = a^{2} a^{b - 2}

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

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

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

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

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

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

数を足す:

1 + 1 = 2

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

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

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

完全平方式を適用する:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = cos^{2} (x)

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

簡素化

1^{2} - 2 \cdot 1 \cdot cos^{2} (x) + (cos^{2} (x))^{2} : 1 - 2 cos^{2} (x) + cos^{4} (x)

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

規則を適用

1^{a} = 1

1^{2} = 1

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

2 \cdot 1 \cdot cos^{2} (x) = 2 cos^{2} (x)

2 \cdot 1 \cdot cos^{2} (x)

数を乗じる:

2 \cdot 1 = 2

= 2 cos^{2} (x)

(cos^{2} (x))^{2} = cos^{4} (x)

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

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= cos^{2 \cdot 2} (x)

数を乗じる:

2 \cdot 2 = 4

= cos^{4} (x)

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

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

cos^{4} (x) + cos^{4} (x) = 2 cos^{4} (x)

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

- 1 + 1 = 0

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} - 2 u^{2} = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

2 v^{2} - 2 v = 0

解く

2 v^{2} - 2 v = 0 : v = 1, v = 0

2 v^{2} - 2 v = 0

解くとthe二次式

2 v^{2} - 2 v = 0

二次Equationの公式:

次の場合:

a = 2, b = - 2, c = 0

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

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

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

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

指数の規則を適用する:

n

が偶数であれば

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

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

= 2^{2} - 4 \cdot 2 \cdot 0

規則を適用

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

, 以下を想定

a \geq 0

= 2

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

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

2 + 2 = 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

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

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

規則を適用

- (- a) = a

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

数を引く:

2 - 2 = 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:

v = 1, v = 0

v = 1, v = 0

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} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

解答は

u = 1, u = - 1, u = 0

代用を戻す

u = cos (x)

cos (x) = 1, cos (x) = - 1, cos (x) = 0

cos (x) = 1, cos (x) = - 1, cos (x) = 0

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

人気の例

2sinh(x)-2cosh(x)-2=0

2 sinh (x) - 2 cosh (x) - 2 = 0

sec^2(x)+cot^2(x)=csc^2(x)

sec^{2} (x) + cot^{2} (x) = csc^{2} (x)

(sin(79)}{12}=\frac{sin(B))/7

\frac{sin ( 7 9 ^{\circ} )}{12} = \frac{sin ( B )}{7}

((sin^2(x)))/((1-cos(x)))=1.23

\frac{( sin ^{2} ( x ) )}{( 1 - cos ( x ) )} = 1.23

-1=tan(2(0)^3+C)

- 1 = tan (2 (0)^{3} + C)