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人気のある三角関数 >

2sin^2(x)-1=0,-pi<= x<= pi

解

2 sin^{2} (x) - 1 = 0, - π \leq x \leq π

解

x = \frac{π}{4}, x = \frac{3 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}

+1

度

x = 4 5^{\circ}, x = 13 5^{\circ}, x = - 13 5^{\circ}, x = - 4 5^{\circ}

解答ステップ

2 sin^{2} (x) - 1 = 0, - π \leq x \leq π

置換で解く

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

仮定:

sin (x) = u

2 u^{2} - 1 = 0

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

2 u^{2} - 1 = 0

1

を右側に移動します

2 u^{2} - 1 = 0

両辺に

1

を足す

2 u^{2} - 1 + 1 = 0 + 1

簡素化

2 u^{2} = 1

2 u^{2} = 1

以下で両辺を割る

2

2 u^{2} = 1

以下で両辺を割る

2

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

簡素化

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

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

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{2}, u = - \frac{1}{2}

代用を戻す

u = sin (x)

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

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

sin (x) = \frac{1}{2}, - π \leq x \leq π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, - π \leq x \leq π

以下の一般解

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

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 = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

範囲の解答

- π \leq x \leq π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, - π \leq x \leq π : x = - \frac{3 π}{4}, x = - \frac{π}{4}

sin (x) = - \frac{1}{2}, - π \leq x \leq π

以下の一般解

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

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 = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

範囲の解答

- π \leq x \leq π

x = - \frac{3 π}{4}, x = - \frac{π}{4}

すべての解を組み合わせる

x = \frac{π}{4}, x = \frac{3 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}

グラフ

人気の例

sin (θ) = - 0.22

3/2 sec(θ)+3cos(θ)=3

\frac{3}{2} sec (θ) + 3 cos (θ) = 3

1 = 7 cos (\frac{π}{3} t)

12cos^2(x)+2cos(x)-2=0

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

sin (θ) = - 0.94