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人気のある三角関数 >

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

解

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

解

x = 2 πn, x = π + 2 πn

+1

度

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

解答ステップ

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

両辺から

cos^{2} (x) - 2 sin^{4} (x)

を引く

sin^{2} (x) + 1 - cos^{2} (x) + 2 sin^{4} (x) = 0

三角関数の公式を使用して書き換える

1 - cos^{2} (x) + sin^{2} (x) + 2 sin^{4} (x)

ピタゴラスの公式を使用する:

1 = cos^{2} (x) + sin^{2} (x)

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

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

簡素化

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

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

置換で解く

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

仮定:

sin (x) = u

2 u^{2} + 2 u^{4} = 0

2 u^{2} + 2 u^{4} = 0 : u = 0, u = i, u = - i

2 u^{2} + 2 u^{4} = 0

標準的な形式で書く

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

2 u^{4} + 2 u^{2} = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

2 v^{2} + 2 v = 0

解く

2 v^{2} + 2 v = 0 : v = 0, v = - 1

2 v^{2} + 2 v = 0

解くとthe二次式

2 v^{2} + 2 v = 0

二次Equationの公式:

次の場合:

a = 2, b = 2, c = 0

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

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

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

2^{2} - 4 \cdot 2 \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

v_{1, 2} = \frac{- 2 \pm 2}{2 \cdot 2}

解を分離する

v_{1} = \frac{- 2 + 2}{2 \cdot 2}, v_{2} = \frac{- 2 - 2}{2 \cdot 2}

v = \frac{- 2 + 2}{2 \cdot 2} : 0

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

数を足す/引く:

- 2 + 2 = 0

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

数を乗じる:

2 \cdot 2 = 4

= \frac{0}{4}

規則を適用

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

= 0

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

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

数を引く:

- 2 - 2 = - 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:

v = 0, v = - 1

v = 0, v = - 1

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

解く

u^{2} = - 1 : u = i, u = - i

u^{2} = - 1

x^{2} = f (a)

の場合, 解は

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

u = - 1, u = - - 1

簡素化

- 1 : i

- 1

虚数の規則を適用する:

- 1 = i

= i

簡素化

- - 1 : - i

- - 1

虚数の規則を適用する:

- 1 = i

= - i

u = i, u = - i

解答は

u = 0, u = i, u = - i

代用を戻す

u = sin (x)

sin (x) = 0, sin (x) = i, sin (x) = - i

sin (x) = 0, sin (x) = i, sin (x) = - i

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) = i :

解なし

sin (x) = i

解なし

sin (x) = - i :

解なし

sin (x) = - i

解なし

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn

グラフ

人気の例

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

cos(2a)= 2/3 ,sin(a)

cos (2 a) = \frac{2}{3}, sin (a)

cot (2 x - 3) - 1 = 0

4sin^{32}(x)-3cos^2(x)=0

4 sin^{32} (x) - 3 cos^{2} (x) = 0

4tan^2(x)+15sec(x)=0

4 tan^{2} (x) + 15 sec (x) = 0