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人気のある三角関数 >

sin(x^2+1)=0

解

sin (x^{2} + 1) = 0

解

x = 2 πn - 1, x = - 2 πn - 1, x = π + 2 πn - 1, x = - π + 2 πn - 1

+1

度

x = 0^{\circ} + 131.69538 \dots^{\circ} n, x = 0^{\circ} - 131.69538 \dots^{\circ} n, x = 0^{\circ} + 166.30368 \dots^{\circ} n, x = 0^{\circ} - 166.30368 \dots^{\circ} n

解答ステップ

sin (x^{2} + 1) = 0

以下の一般解

sin (x^{2} + 1) = 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^{2} + 1 = 0 + 2 πn, x^{2} + 1 = π + 2 πn

x^{2} + 1 = 0 + 2 πn, x^{2} + 1 = π + 2 πn

解く

x^{2} + 1 = 0 + 2 πn : x = 2 πn - 1, x = - 2 πn - 1

x^{2} + 1 = 0 + 2 πn

1

を右側に移動します

x^{2} + 1 = 0 + 2 πn

両辺から

1

を引く

x^{2} + 1 - 1 = 0 + 2 πn - 1

簡素化

x^{2} = 2 πn - 1

x^{2} = 2 πn - 1

x^{2} = f (a)

の場合, 解は

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

x = 2 πn - 1, x = - 2 πn - 1

解く

x^{2} + 1 = π + 2 πn : x = π + 2 πn - 1, x = - π + 2 πn - 1

x^{2} + 1 = π + 2 πn

1

を右側に移動します

x^{2} + 1 = π + 2 πn

両辺から

1

を引く

x^{2} + 1 - 1 = π + 2 πn - 1

簡素化

x^{2} = π + 2 πn - 1

x^{2} = π + 2 πn - 1

x^{2} = f (a)

の場合, 解は

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

x = π + 2 πn - 1, x = - π + 2 πn - 1

x = 2 πn - 1, x = - 2 πn - 1, x = π + 2 πn - 1, x = - π + 2 πn - 1

グラフ

人気の例

cos(2x)+sin^2(x)=0

cos (2 x) + sin^{2} (x) = 0

1+cos(x)=sqrt(3)sin(x)

1 + cos (x) = 3 sin (x)

3 sin (x) = 0

1+sin(x)=cos(x)

1 + sin (x) = cos (x)

sin(x+pi/2)-cos^2(x)=0

sin (x + \frac{π}{2}) - cos^{2} (x) = 0