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人気のある三角関数 >

sqrt(2)sin^2(θ)+sin(θ)=0

解

2 sin^{2} (θ) + sin (θ) = 0

解

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

+1

度

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 22 5^{\circ} + 36 0^{\circ} n, θ = 31 5^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin^{2} (θ) + sin (θ) = 0

置換で解く

2 sin^{2} (θ) + sin (θ) = 0

仮定:

sin (θ) = u

2 u^{2} + u = 0

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

2 u^{2} + u = 0

解くとthe二次式

2 u^{2} + u = 0

二次Equationの公式:

次の場合:

a = 2, b = 1, c = 0

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

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

1^{2} - 42 \cdot 0 = 1

1^{2} - 42 \cdot 0

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 0 \cdot 2

規則を適用

0 \cdot a = 0

= 1 - 0

数を引く:

1 - 0 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

数を足す/引く:

- 1 + 1 = 0

= \frac{0}{2 2}

規則を適用

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

= 0

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

\frac{- 1 - 1}{2 2}

数を引く:

- 1 - 1 = - 2

= \frac{- 2}{2 2}

分数の規則を適用する:

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

= - \frac{2}{2 2}

数を割る:

\frac{2}{2} = 1

= - \frac{1}{2}

有理化する

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

共役で乗じる

\frac{2}{2}

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

1 \cdot 2 = 2

22 = 2

22

累乗根の規則を適用する:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

二次equationの解:

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

代用を戻す

u = sin (θ)

sin (θ) = 0, sin (θ) = - \frac{2}{2}

sin (θ) = 0, sin (θ) = - \frac{2}{2}

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

以下の一般解

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

解く

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

sin (θ) = - \frac{2}{2} : θ = \frac{5 π}{4} + 2 πn, θ = \frac{7 π}{4} + 2 πn

sin (θ) = - \frac{2}{2}

以下の一般解

sin (θ) = - \frac{2}{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}

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

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

すべての解を組み合わせる

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

グラフ

人気の例

tan^2(x)=1-sec^2(x)

tan^{2} (x) = 1 - sec^{2} (x)

sec (x) = cos (x)

cot(2x)=(sqrt(3))/3

cot (2 x) = \frac{3}{3}

cos(2θ)+cos(θ)=2

cos (2 θ) + cos (θ) = 2

4 sin (x) = 0