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人気のある三角関数 >

(tan(x)+cos(x))/((1+sin(x)))= 1/(cos(x))

解

\frac{tan ( x ) + cos ( x )}{( 1 + sin ( x ) )} = \frac{1}{cos ( 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

\frac{tan ( x ) + cos ( x )}{( 1 + sin ( x ) )} = \frac{1}{cos ( x )}

両辺から

\frac{1}{cos ( x )}

を引く

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - \frac{1}{cos ( x )} = 0

簡素化

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - \frac{1}{cos ( x )} : \frac{cos ( x ) tan ( x ) + cos ^{2} ( x ) - sin ( x ) - 1}{cos ( x ) ( sin ( x ) + 1 )}

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} - \frac{1}{cos ( x )}

以下の最小公倍数:

1 + sin (x), cos (x) : cos (x) (sin (x) + 1)

1 + sin (x), cos (x)

最小公倍数 (LCM)

1 + sin (x)

または以下のいずれかに現れる因数で構成された式を計算する:

cos (x)

= cos (x) (sin (x) + 1)

LCMに基づいて分数を調整する

該当する分母を乗じてLCMに変えるために

必要な量で各分子を乗じる

cos (x) (sin (x) + 1)

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )}

の場合

:

分母と分子に以下を乗じる:

cos (x)

\frac{tan ( x ) + cos ( x )}{1 + sin ( x )} = \frac{( tan ( x ) + cos ( x ) ) cos ( x )}{( 1 + sin ( x ) ) cos ( x )}

\frac{1}{cos ( x )}

の場合

:

分母と分子に以下を乗じる:

sin (x) + 1

\frac{1}{cos ( x )} = \frac{1 \cdot ( sin ( x ) + 1 )}{cos ( x ) ( sin ( x ) + 1 )} = \frac{sin ( x ) + 1}{cos ( x ) ( sin ( x ) + 1 )}

= \frac{( tan ( x ) + cos ( x ) ) cos ( x )}{( 1 + sin ( x ) ) cos ( x )} - \frac{sin ( x ) + 1}{cos ( x ) ( sin ( x ) + 1 )}

分母が等しいので, 分数を組み合わせる:

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

= \frac{( tan ( x ) + cos ( x ) ) cos ( x ) - ( sin ( x ) + 1 )}{cos ( x ) ( sin ( x ) + 1 )}

拡張

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

(tan (x) + cos (x)) cos (x) - (sin (x) + 1)

= cos (x) (tan (x) + cos (x)) - (sin (x) + 1)

拡張

cos (x) (tan (x) + cos (x)) : cos (x) tan (x) + cos^{2} (x)

cos (x) (tan (x) + cos (x))

分配法則を適用する:

a (b + c) = ab + a c

a = cos (x), b = tan (x), c = cos (x)

= cos (x) tan (x) + cos (x) cos (x)

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

cos (x) cos (x)

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

数を足す:

1 + 1 = 2

= cos^{2} (x)

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

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

- (sin (x) + 1) : - sin (x) - 1

- (sin (x) + 1)

括弧を分配する

= - (sin (x)) - (1)

マイナス・プラスの規則を適用する

+ (- a) = - a

= - sin (x) - 1

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

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

\frac{cos ( x ) tan ( x ) + cos ^{2} ( x ) - sin ( x ) - 1}{cos ( x ) ( sin ( x ) + 1 )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

三角関数の公式を使用して書き換える

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

基本的な三角関数の公式を使用する:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

簡素化

- 1 + cos^{2} (x) - sin (x) + cos (x) \frac{sin ( x )}{cos ( x )} : - 1 + cos^{2} (x)

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

cos (x) \frac{sin ( x )}{cos ( x )} = sin (x)

cos (x) \frac{sin ( x )}{cos ( x )}

分数を乗じる:

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

= \frac{sin ( x ) cos ( x )}{cos ( x )}

共通因数を約分する:

cos (x)

= sin (x)

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

類似した元を足す:

- sin (x) + sin (x) = 0

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

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

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

置換で解く

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

仮定:

cos (x) = u

- 1 + u^{2} = 0

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

- 1 + u^{2} = 0

1

を右側に移動します

- 1 + u^{2} = 0

両辺に

1

を足す

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

簡素化

u^{2} = 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

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

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

代用を戻す

u = cos (x)

cos (x) = 1, cos (x) = - 1

cos (x) = 1, cos (x) = - 1

cos (x) = 1 : x = 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 = 0 + 2 πn

x = 0 + 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

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 = 2 πn, x = π + 2 πn

グラフ

人気の例

sin(θ)= 4/5 ,0<= θ<= pi/2

sin (θ) = \frac{4}{5}, 0 \leq θ \leq \frac{π}{2}

((tan(x)+2sin(x)))/((tan(x)-2sin(x)))=3

\frac{( tan ( x ) + 2 sin ( x ) )}{( tan ( x ) - 2 sin ( x ) )} = 3

1 - sin (2 x) = 0

2cos(3x)+cos(2x)+1=0

2 cos (3 x) + cos (2 x) + 1 = 0

sin(x)=sin(pi/5)

sin (x) = sin (\frac{π}{5})