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 >

cos(2x)= 3/4-5cos^2(x)

解

cos (2 x) = \frac{3}{4} - 5 cos^{2} (x)

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

cos (2 x) = \frac{3}{4} - 5 cos^{2} (x)

両辺から

\frac{3}{4} - 5 cos^{2} (x)

を引く

cos (2 x) - \frac{3}{4} + 5 cos^{2} (x) = 0

簡素化

cos (2 x) - \frac{3}{4} + 5 cos^{2} (x) : \frac{4 cos ( 2 x ) - 3 + 20 cos ^{2} ( x )}{4}

cos (2 x) - \frac{3}{4} + 5 cos^{2} (x)

元を分数に変換する:

cos (2 x) = \frac{cos ( 2 x ) 4}{4}, 5 cos^{2} (x) = \frac{5 cos ^{2} ( x ) 4}{4}

= \frac{cos ( 2 x ) \cdot 4}{4} - \frac{3}{4} + \frac{5 cos ^{2} ( x ) \cdot 4}{4}

分母が等しいので, 分数を組み合わせる:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{cos ( 2 x ) \cdot 4 - 3 + 5 cos ^{2} ( x ) \cdot 4}{4}

数を乗じる:

5 \cdot 4 = 20

= \frac{4 cos ( 2 x ) - 3 + 20 cos ^{2} ( x )}{4}

\frac{4 cos ( 2 x ) - 3 + 20 cos ^{2} ( x )}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 cos (2 x) - 3 + 20 cos^{2} (x) = 0

三角関数の公式を使用して書き換える

- 3 + 20 cos^{2} (x) + 4 cos (2 x)

2倍角の公式を使用:

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

= - 3 + 20 cos^{2} (x) + 4 (2 cos^{2} (x) - 1)

簡素化

- 3 + 20 cos^{2} (x) + 4 (2 cos^{2} (x) - 1) : 28 cos^{2} (x) - 7

- 3 + 20 cos^{2} (x) + 4 (2 cos^{2} (x) - 1)

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 2 cos^{2} (x), c = 1

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

簡素化

4 \cdot 2 cos^{2} (x) - 4 \cdot 1 : 8 cos^{2} (x) - 4

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

数を乗じる:

4 \cdot 2 = 8

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

数を乗じる:

4 \cdot 1 = 4

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

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

= - 3 + 20 cos^{2} (x) + 8 cos^{2} (x) - 4

簡素化

- 3 + 20 cos^{2} (x) + 8 cos^{2} (x) - 4 : 28 cos^{2} (x) - 7

- 3 + 20 cos^{2} (x) + 8 cos^{2} (x) - 4

類似した元を足す:

20 cos^{2} (x) + 8 cos^{2} (x) = 28 cos^{2} (x)

= - 3 + 28 cos^{2} (x) - 4

条件のようなグループ

= 28 cos^{2} (x) - 3 - 4

数を引く:

- 3 - 4 = - 7

= 28 cos^{2} (x) - 7

= 28 cos^{2} (x) - 7

= 28 cos^{2} (x) - 7

- 7 + 28 cos^{2} (x) = 0

置換で解く

- 7 + 28 cos^{2} (x) = 0

仮定:

cos (x) = u

- 7 + 28 u^{2} = 0

- 7 + 28 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

- 7 + 28 u^{2} = 0

7

を右側に移動します

- 7 + 28 u^{2} = 0

両辺に

7

を足す

- 7 + 28 u^{2} + 7 = 0 + 7

簡素化

28 u^{2} = 7

28 u^{2} = 7

以下で両辺を割る

28

28 u^{2} = 7

以下で両辺を割る

28

\frac{28 u ^{2}}{28} = \frac{7}{28}

簡素化

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

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}

代用を戻す

u = cos (x)

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

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

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

cos (x) = \frac{1}{2}

以下の一般解

cos (x) = \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}

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

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

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

以下の一般解

cos (x) = - \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}

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

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

sin(x+pi/9)=(sqrt(3))/2

sin (x + \frac{π}{9}) = \frac{3}{2}

3cos(2θ)=3sin(θ)

3 cos (2 θ) = 3 sin (θ)

-sec(θ)cot(θ)-sec(θ)=-2sec(θ)

- sec (θ) cot (θ) - sec (θ) = - 2 sec (θ)

sin(5x-27)=cos(6x+13.6)

sin (5 x - 27) = cos (6 x + 13.6)

sin(30-x)+cos(x)=0

sin (3 0^{\circ} - x) + cos (x) = 0