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人気のある三角関数 >

4sin^2(x)+2sin(x)=0

解

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

解

x = 2 πn, x = π + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

sin (x) = u

4 u^{2} + 2 u = 0

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

4 u^{2} + 2 u = 0

解くとthe二次式

4 u^{2} + 2 u = 0

二次Equationの公式:

次の場合:

a = 4, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

2^{2} - 4 \cdot 4 \cdot 0 = 2

2^{2} - 4 \cdot 4 \cdot 0

規則を適用

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

, 以下を想定

a \geq 0

= 2

u_{1, 2} = \frac{- 2 \pm 2}{2 \cdot 4}

解を分離する

u_{1} = \frac{- 2 + 2}{2 \cdot 4}, u_{2} = \frac{- 2 - 2}{2 \cdot 4}

u = \frac{- 2 + 2}{2 \cdot 4} : 0

\frac{- 2 + 2}{2 \cdot 4}

数を足す/引く:

- 2 + 2 = 0

= \frac{0}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{0}{8}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 2 - 2}{2 \cdot 4} : - \frac{1}{2}

\frac{- 2 - 2}{2 \cdot 4}

数を引く:

- 2 - 2 = - 4

= \frac{- 4}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 4}{8}

分数の規則を適用する:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{8}

共通因数を約分する:

4

= - \frac{1}{2}

二次equationの解:

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

代用を戻す

u = sin (x)

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

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

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

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

グラフ

人気の例

tan(x)=sqrt(3),-pi/3 <= x<= (3pi)/2

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