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人気のある三角関数 >

([sin(x)+cos(x)])/(cos(x))=sin(x)+1

解

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

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{[ sin ( x ) + cos ( x ) ]}{cos ( x )} = sin (x) + 1

両辺から

sin (x) + 1

を引く

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

簡素化

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

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

元を分数に変換する:

sin (x) = \frac{sin ( x ) cos ( x )}{cos ( x )}, 1 = \frac{1 cos ( x )}{cos ( x )}

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

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{sin ( x ) + cos ( x ) - sin ( x ) cos ( x ) - 1 \cdot cos ( x )}{cos ( x )}

類似した元を足す:

cos (x) - 1 \cdot cos (x) = 0

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

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

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

sin (x) - sin (x) cos (x) = 0

因数

sin (x) - sin (x) cos (x) : - sin (x) (cos (x) - 1)

sin (x) - sin (x) cos (x)

共通項をくくり出す

- sin (x)

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

- sin (x) (cos (x) - 1) = 0

各部分を別個に解く

sin (x) = 0 or cos (x) - 1 = 0

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

cos (x) - 1 = 0 : x = 2 πn

cos (x) - 1 = 0

1

を右側に移動します

cos (x) - 1 = 0

両辺に

1

を足す

cos (x) - 1 + 1 = 0 + 1

簡素化

cos (x) = 1

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

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn

グラフ

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