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人気のある三角関数 >

-2sin(x)+5sin^2(x)=0

解

- 2 sin (x) + 5 sin^{2} (x) = 0

解

x = 0.41151 \dots + 2 πn, x = π - 0.41151 \dots + 2 πn, x = 2 πn, x = π + 2 πn

+1

度

x = 23.57817 \dots^{\circ} + 36 0^{\circ} n, x = 156.42182 \dots^{\circ} + 36 0^{\circ} n, x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

- 2 sin (x) + 5 sin^{2} (x) = 0

置換で解く

- 2 sin (x) + 5 sin^{2} (x) = 0

仮定:

sin (x) = u

- 2 u + 5 u^{2} = 0

- 2 u + 5 u^{2} = 0 : u = \frac{2}{5}, u = 0

- 2 u + 5 u^{2} = 0

標準的な形式で書く

a x^{2} + b x + c = 0

5 u^{2} - 2 u = 0

解くとthe二次式

5 u^{2} - 2 u = 0

二次Equationの公式:

次の場合:

a = 5, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

(- 2)^{2} - 4 \cdot 5 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 5 \cdot 0

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 2)^{2} = 2^{2}

= 2^{2} - 4 \cdot 5 \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:

n a^{n} = a,

, 以下を想定

a \geq 0

= 2

u_{1, 2} = \frac{- ( - 2 ) \pm 2}{2 \cdot 5}

解を分離する

u_{1} = \frac{- ( - 2 ) + 2}{2 \cdot 5}, u_{2} = \frac{- ( - 2 ) - 2}{2 \cdot 5}

u = \frac{- ( - 2 ) + 2}{2 \cdot 5} : \frac{2}{5}

\frac{- ( - 2 ) + 2}{2 \cdot 5}

規則を適用

- (- a) = a

= \frac{2 + 2}{2 \cdot 5}

数を足す:

2 + 2 = 4

= \frac{4}{2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{4}{10}

共通因数を約分する:

2

= \frac{2}{5}

u = \frac{- ( - 2 ) - 2}{2 \cdot 5} : 0

\frac{- ( - 2 ) - 2}{2 \cdot 5}

規則を適用

- (- a) = a

= \frac{2 - 2}{2 \cdot 5}

数を引く:

2 - 2 = 0

= \frac{0}{2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{0}{10}

規則を適用

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

= 0

二次equationの解:

u = \frac{2}{5}, u = 0

代用を戻す

u = sin (x)

sin (x) = \frac{2}{5}, sin (x) = 0

sin (x) = \frac{2}{5}, sin (x) = 0

sin (x) = \frac{2}{5} : x = arcsin (\frac{2}{5}) + 2 πn, x = π - arcsin (\frac{2}{5}) + 2 πn

sin (x) = \frac{2}{5}

三角関数の逆数プロパティを適用する

sin (x) = \frac{2}{5}

以下の一般解

sin (x) = \frac{2}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{2}{5}) + 2 πn, x = π - arcsin (\frac{2}{5}) + 2 πn

x = arcsin (\frac{2}{5}) + 2 πn, x = π - arcsin (\frac{2}{5}) + 2 πn

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

すべての解を組み合わせる

x = arcsin (\frac{2}{5}) + 2 πn, x = π - arcsin (\frac{2}{5}) + 2 πn, x = 2 πn, x = π + 2 πn

10進法形式で解を証明する

x = 0.41151 \dots + 2 πn, x = π - 0.41151 \dots + 2 πn, x = 2 πn, x = π + 2 πn

グラフ

人気の例

sqrt(3)*tan^2(x)-1=0

3 \cdot tan^{2} (x) - 1 = 0

cos^4(x)=(sin^2(x)-1)/4

cos^{4} (x) = \frac{sin ^{2} ( x ) - 1}{4}

sqrt(2)*sin(x)+1=0

2 \cdot sin (x) + 1 = 0

5cos^2(2x)+4cos^2(x)-5=0

5 cos^{2} (2 x) + 4 cos^{2} (x) - 5 = 0

sin(3x+10)=cos(x+24)

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