ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

sec^4(x)-4sec^2(x)=0

解

sec^{4} (x) - 4 sec^{2} (x) = 0

解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

解答ステップ

sec^{4} (x) - 4 sec^{2} (x) = 0

置換で解く

sec^{4} (x) - 4 sec^{2} (x) = 0

仮定:

sec (x) = u

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

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

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

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

v^{2} - 4 v = 0

解く

v^{2} - 4 v = 0 : v = 4, v = 0

v^{2} - 4 v = 0

解くとthe二次式

v^{2} - 4 v = 0

二次Equationの公式:

次の場合:

a = 1, b = - 4, c = 0

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

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

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

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

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 1 \cdot 0

規則を適用

0 \cdot a = 0

= 4^{2} - 0

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

= 4^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 4

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

解を分離する

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

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

\frac{- ( - 4 ) + 4}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{4 + 4}{2 \cdot 1}

数を足す:

4 + 4 = 8

= \frac{8}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{8}{2}

数を割る:

\frac{8}{2} = 4

= 4

v = \frac{- ( - 4 ) - 4}{2 \cdot 1} : 0

\frac{- ( - 4 ) - 4}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{4 - 4}{2 \cdot 1}

数を引く:

4 - 4 = 0

= \frac{0}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{0}{2}

規則を適用

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

= 0

二次equationの解:

v = 4, v = 0

v = 4, v = 0

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 4 : u = 2, u = - 2

u^{2} = 4

x^{2} = f (a)

の場合, 解は

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

u = 4, u = - 4

4 = 2

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

- 4 = - 2

- 4

4 = 2

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

= - 2

u = 2, u = - 2

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 = 2, u = - 2, u = 0

代用を戻す

u = sec (x)

sec (x) = 2, sec (x) = - 2, sec (x) = 0

sec (x) = 2, sec (x) = - 2, sec (x) = 0

sec (x) = 2 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sec (x) = 2

以下の一般解

sec (x) = 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

sec (x) = - 2 : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sec (x) = - 2

以下の一般解

sec (x) = - 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sec (x) = 0 :

解なし

sec (x) = 0

sec (x) \leq - 1 or sec (x) \geq 1

解なし

すべての解を組み合わせる

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

グラフ

人気の例

(cos(x))/(sec(x))-(sin(x))/(csc(x))=1

\frac{cos ( x )}{sec ( x )} - \frac{sin ( x )}{csc ( x )} = 1

tan (x) = \frac{12}{35}

4 = sec (θ)

1+sqrt(2)sin(x)=cos(2x)

1 + 2 sin (x) = cos (2 x)

12-sin(θ)=cos(2θ)

12 - sin (θ) = cos (2 θ)