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 >

2sin^2(t)cos(t)=0

解

2 sin^{2} (t) cos (t) = 0

解

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

+1

度

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n, t = 9 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin^{2} (t) cos (t) = 0

各部分を別個に解く

sin^{2} (t) = 0 or cos (t) = 0

sin^{2} (t) = 0 : t = 2 πn, t = π + 2 πn

sin^{2} (t) = 0

規則を適用

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

sin (t) = 0

以下の一般解

sin (t) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

解く

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

以下の一般解

cos (t) = 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}

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

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

すべての解を組み合わせる

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

グラフ

人気の例

5sin(x)-12cos(x)=13

5 sin (x) - 12 cos (x) = 13

6 sin (θ) = 1

2cos((3x-pi)/6)+1=0,0<x<2pi

2 cos (\frac{3 x - π}{6}) + 1 = 0, 0 < x < 2 π

2cos^2(a)+cos(a)-1=0

2 cos^{2} (a) + cos (a) - 1 = 0

arctan(e^x)= pi/4

arctan (e^{x}) = \frac{π}{4}