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人気のある三角関数 >

2sin^2(t)+3sin(t)+1=0

解

2 sin^{2} (t) + 3 sin (t) + 1 = 0

解

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn

+1

度

t = 21 0^{\circ} + 36 0^{\circ} n, t = 33 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin^{2} (t) + 3 sin (t) + 1 = 0

置換で解く

2 sin^{2} (t) + 3 sin (t) + 1 = 0

仮定:

sin (t) = u

2 u^{2} + 3 u + 1 = 0

2 u^{2} + 3 u + 1 = 0 : u = - \frac{1}{2}, u = - 1

2 u^{2} + 3 u + 1 = 0

解くとthe二次式

2 u^{2} + 3 u + 1 = 0

二次Equationの公式:

次の場合:

a = 2, b = 3, c = 1

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 \cdot 1}{2 \cdot 2}

3^{2} - 4 \cdot 2 \cdot 1 = 1

3^{2} - 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数を引く:

9 - 8 = 1

= 1

規則を適用

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 3 + 1}{2 \cdot 2}, u_{2} = \frac{- 3 - 1}{2 \cdot 2}

u = \frac{- 3 + 1}{2 \cdot 2} : - \frac{1}{2}

\frac{- 3 + 1}{2 \cdot 2}

数を足す/引く:

- 3 + 1 = - 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{4}

分数の規則を適用する:

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

= - \frac{2}{4}

共通因数を約分する:

2

= - \frac{1}{2}

u = \frac{- 3 - 1}{2 \cdot 2} : - 1

\frac{- 3 - 1}{2 \cdot 2}

数を引く:

- 3 - 1 = - 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{4}

分数の規則を適用する:

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

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

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

代用を戻す

u = sin (t)

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

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

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

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

以下の一般解

sin (t) = - \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}

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

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

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

sin (t) = - 1

以下の一般解

sin (t) = - 1

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 = \frac{3 π}{2} + 2 πn

t = \frac{3 π}{2} + 2 πn

すべての解を組み合わせる

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{3 π}{2} + 2 πn

グラフ

人気の例

2sin^2(θ)-7sin(θ)-4=0

2 sin^{2} (θ) - 7 sin (θ) - 4 = 0

cos (4 θ) = 1

tan(2x)+sec(2x)=3

tan (2 x) + sec (2 x) = 3

6sin(x)-5cos(x)=7

6 sin (x) - 5 cos (x) = 7

3sin^2(x)-7sin(x)=6

3 sin^{2} (x) - 7 sin (x) = 6