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人気のある三角関数 >

4cos^4(θ)-5cos^2(θ)+1=0

解

4 cos^{4} (θ) - 5 cos^{2} (θ) + 1 = 0

解

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

+1

度

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 6 0^{\circ} + 36 0^{\circ} n, θ = 30 0^{\circ} + 36 0^{\circ} n, θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

解答ステップ

4 cos^{4} (θ) - 5 cos^{2} (θ) + 1 = 0

置換で解く

4 cos^{4} (θ) - 5 cos^{2} (θ) + 1 = 0

仮定:

cos (θ) = u

4 u^{4} - 5 u^{2} + 1 = 0

4 u^{4} - 5 u^{2} + 1 = 0 : u = 1, u = - 1, u = \frac{1}{2}, u = - \frac{1}{2}

4 u^{4} - 5 u^{2} + 1 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

4 v^{2} - 5 v + 1 = 0

解く

4 v^{2} - 5 v + 1 = 0 : v = 1, v = \frac{1}{4}

4 v^{2} - 5 v + 1 = 0

解くとthe二次式

4 v^{2} - 5 v + 1 = 0

二次Equationの公式:

次の場合:

a = 4, b = - 5, c = 1

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

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

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

指数の規則を適用する:

n

が偶数であれば

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

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 4 \cdot 1

数を乗じる:

4 \cdot 4 \cdot 1 = 16

= 5^{2} - 16

5^{2} = 25

= 25 - 16

数を引く:

25 - 16 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

v_{1, 2} = \frac{- ( - 5 ) \pm 3}{2 \cdot 4}

解を分離する

v_{1} = \frac{- ( - 5 ) + 3}{2 \cdot 4}, v_{2} = \frac{- ( - 5 ) - 3}{2 \cdot 4}

v = \frac{- ( - 5 ) + 3}{2 \cdot 4} : 1

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

規則を適用

- (- a) = a

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

数を足す:

5 + 3 = 8

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

数を乗じる:

2 \cdot 4 = 8

= \frac{8}{8}

規則を適用

\frac{a}{a} = 1

= 1

v = \frac{- ( - 5 ) - 3}{2 \cdot 4} : \frac{1}{4}

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

規則を適用

- (- a) = a

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

数を引く:

5 - 3 = 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}

二次equationの解:

v = 1, v = \frac{1}{4}

v = 1, v = \frac{1}{4}

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}{4} : u = \frac{1}{2}, u = - \frac{1}{2}

u^{2} = \frac{1}{4}

x^{2} = f (a)

の場合, 解は

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

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

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

\frac{1}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{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

= \frac{1}{2}

規則を適用

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

簡素化

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{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

= \frac{1}{2}

= - \frac{1}{2}

規則を適用

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

解答は

u = 1, u = - 1, u = \frac{1}{2}, u = - \frac{1}{2}

代用を戻す

u = cos (θ)

cos (θ) = 1, cos (θ) = - 1, cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = 1, cos (θ) = - 1, cos (θ) = \frac{1}{2}, cos (θ) = - \frac{1}{2}

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

以下の一般解

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

解く

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

以下の一般解

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

cos (θ) = \frac{1}{2} : θ = \frac{π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

cos (θ) = \frac{1}{2}

以下の一般解

cos (θ) = \frac{1}{2}

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}

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

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

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

以下の一般解

cos (θ) = - \frac{1}{2}

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}

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

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

すべての解を組み合わせる

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

グラフ

人気の例

solvefor x,cos(x)=sin(2x)

so l v e f or x, cos (x) = sin (2 x)

sin(2t)-sin(t)=0

sin (2 t) - sin (t) = 0

sin (X) = 0

3 sec (x) = - 6

solvefor x,cos(x+xy)=2

so l v e f or x, cos (x + x y) = 2