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^3(x)=cos(x),0<= x<2pi

解

cos^{3} (x) = cos (x), 0 \leq x < 2 π

解

x = \frac{π}{2}, x = \frac{3 π}{2}, x = π, x = 0

+1

度

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 18 0^{\circ}, x = 0^{\circ}

解答ステップ

cos^{3} (x) = cos (x), 0 \leq x < 2 π

置換で解く

cos^{3} (x) = cos (x)

仮定:

cos (x) = u

u^{3} = u

u^{3} = u : u = 0, u = - 1, u = 1

u^{3} = u

u

を左側に移動します

u^{3} = u

両辺から

u

を引く

u^{3} - u = u - u

簡素化

u^{3} - u = 0

u^{3} - u = 0

因数

u^{3} - u : u (u + 1) (u - 1)

u^{3} - u

共通項をくくり出す

u : u (u^{2} - 1)

u^{3} - u

指数の規則を適用する:

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

u^{3} = u^{2} u

= u^{2} u - u

共通項をくくり出す

u

= u (u^{2} - 1)

= u (u^{2} - 1)

因数

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

1

を書き換え

1^{2}

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

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= u (u + 1) (u - 1)

u (u + 1) (u - 1) = 0

零因子の原則を使用:

ab = 0

ならば

a = 0

または

b = 0

u = 0 or u + 1 = 0 or u - 1 = 0

解く

u + 1 = 0 : u = - 1

u + 1 = 0

1

を右側に移動します

u + 1 = 0

両辺から

1

を引く

u + 1 - 1 = 0 - 1

簡素化

u = - 1

u = - 1

解く

u - 1 = 0 : u = 1

u - 1 = 0

1

を右側に移動します

u - 1 = 0

両辺に

1

を足す

u - 1 + 1 = 0 + 1

簡素化

u = 1

u = 1

解答は

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

代用を戻す

u = cos (x)

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

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

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

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

範囲の解答

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = - 1, 0 \leq x < 2 π : x = π

cos (x) = - 1, 0 \leq x < 2 π

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

範囲の解答

0 \leq x < 2 π

x = π

cos (x) = 1, 0 \leq x < 2 π : x = 0

cos (x) = 1, 0 \leq x < 2 π

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

範囲の解答

0 \leq x < 2 π

x = 0

すべての解を組み合わせる

x = \frac{π}{2}, x = \frac{3 π}{2}, x = π, x = 0

グラフ

人気の例

sin (r) = 0.58

sec (α) = \frac{9}{5}

cos^2(x)+sin(x)= 5/4

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

(8.87)/(9.8)=sin(α)-0.1sqrt(1-sin^2(α))

\frac{8.87}{9.8} = sin (α) - 0.1 1 - sin^{2} (α)

sqrt(2)=csc(x)cot(x)

2 = csc (x) cot (x)