ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

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

解

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

解

x = π + 2 πn

+1

度

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

解答ステップ

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

両辺から

sin^{2} (x)

を引く

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

- 1 + d^{2} - 2 cos (x) + d^{2} cos (x) - 1 : d^{2} cos (x) - 2 cos (x) + d^{2} - 2

- 1 + d^{2} - 2 cos (x) + d^{2} cos (x) - 1

条件のようなグループ

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

数を引く:

- 1 - 1 = - 2

= d^{2} cos (x) - 2 cos (x) + d^{2} - 2

= d^{2} cos (x) - 2 cos (x) + d^{2} - 2

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

置換で解く

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

仮定:

cos (x) = u

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

- 2 + d^{2} - 2 u + u d^{2} = 0 : u = - 1; d \neq = 2, d \neq = - 2

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

d^{2}

を右側に移動します

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

両辺から

d^{2}

を引く

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

簡素化

- 2 - 2 u + u d^{2} = - d^{2}

- 2 - 2 u + u d^{2} = - d^{2}

2

を右側に移動します

- 2 - 2 u + u d^{2} = - d^{2}

両辺に

2

を足す

- 2 - 2 u + u d^{2} + 2 = - d^{2} + 2

簡素化

- 2 u + u d^{2} = - d^{2} + 2

- 2 u + u d^{2} = - d^{2} + 2

因数

- 2 u + u d^{2} : u (- 2 + d^{2})

- 2 u + u d^{2}

共通項をくくり出す

u

= u (- 2 + d^{2})

u (- 2 + d^{2}) = - d^{2} + 2

以下で両辺を割る

- 2 + d^{2}; d \neq = 2, d \neq = - 2

u (- 2 + d^{2}) = - d^{2} + 2

以下で両辺を割る

- 2 + d^{2}; d \neq = 2, d \neq = - 2

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}} = - \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}}; d \neq = 2, d \neq = - 2

簡素化

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}} = - \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}}

簡素化

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}} : u

\frac{u ( - 2 + d ^{2} )}{- 2 + d ^{2}}

共通因数を約分する:

- 2 + d^{2}

= u

簡素化

- \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}} : - 1

- \frac{d ^{2}}{- 2 + d ^{2}} + \frac{2}{- 2 + d ^{2}}

規則を適用

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- d ^{2} + 2}{- 2 + d ^{2}}

くくり出す

- 1

= \frac{- ( d ^{2} - 2 )}{d ^{2} - 2}

改良

= - \frac{d ^{2} - 2}{d ^{2} - 2}

規則を適用

\frac{a}{a} = 1

= - 1

u = - 1; d \neq = 2, d \neq = - 2

u = - 1; d \neq = 2, d \neq = - 2

u = - 1; d \neq = 2, d \neq = - 2

代用を戻す

u = cos (x)

cos (x) = - 1; d \neq = 2, d \neq = - 2

cos (x) = - 1; d \neq = 2, d \neq = - 2

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

以下の一般解

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

すべての解を組み合わせる

x = π + 2 πn

グラフ

人気の例

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

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

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

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

(sin^3(x))/(sin(x))=0

\frac{sin ^{3} ( x )}{sin ( x )} = 0

sin(135-x)=sin(x)

sin (135 - x) = sin (x)

tan (x) - 4 = - 1