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人気のある三角関数 >

(8cos(x)-4)(sin(x))=0,0<= x<2pi

解

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

解

x = \frac{π}{3}, x = \frac{5 π}{3}, x = 0, x = π

+1

度

x = 6 0^{\circ}, x = 30 0^{\circ}, x = 0^{\circ}, x = 18 0^{\circ}

解答ステップ

(8 cos (x) - 4) (sin (x)) = 0, 0 \leq x < 2 π

各部分を別個に解く

8 cos (x) - 4 = 0 or sin (x) = 0

8 cos (x) - 4 = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

8 cos (x) - 4 = 0, 0 \leq x < 2 π

4

を右側に移動します

8 cos (x) - 4 = 0

両辺に

4

を足す

8 cos (x) - 4 + 4 = 0 + 4

簡素化

8 cos (x) = 4

8 cos (x) = 4

以下で両辺を割る

8

8 cos (x) = 4

以下で両辺を割る

8

\frac{8 cos ( x )}{8} = \frac{4}{8}

簡素化

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

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

範囲の解答

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

以下の一般解

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

範囲の解答

0 \leq x < 2 π

x = 0, x = π

すべての解を組み合わせる

x = \frac{π}{3}, x = \frac{5 π}{3}, x = 0, x = π

グラフ

人気の例

sin (x) = 3511

(10)/(sin(30))= 3/(sin(b))

\frac{10}{sin ( 3 0 ^{\circ} )} = \frac{3}{sin ( b )}

tan(θ/2-pi/6)=-1,0<= θ<2pi

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

arctan(x+2)-arctan(x+1)= pi/4

arctan (x + 2) - arctan (x + 1) = \frac{π}{4}

3 cos (2 x) = 1