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 >

tan(x)=5sin(x)

解

tan (x) = 5 sin (x)

解

x = 2 πn, x = π + 2 πn, x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn

+1

度

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 78.46304 \dots^{\circ} + 36 0^{\circ} n, x = 281.53695 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

tan (x) = 5 sin (x)

両辺から

5 sin (x)

を引く

tan (x) - 5 sin (x) = 0

サイン, コサインで表わす

tan (x) - 5 sin (x)

基本的な三角関数の公式を使用する:

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

= \frac{sin ( x )}{cos ( x )} - 5 sin (x)

簡素化

\frac{sin ( x )}{cos ( x )} - 5 sin (x) : \frac{sin ( x ) - 5 sin ( x ) cos ( x )}{cos ( x )}

\frac{sin ( x )}{cos ( x )} - 5 sin (x)

元を分数に変換する:

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

= \frac{sin ( x )}{cos ( x )} - \frac{5 sin ( x ) 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 ) - 5 sin ( x ) cos ( x )}{cos ( x )}

= \frac{sin ( x ) - 5 sin ( x ) cos ( x )}{cos ( x )}

\frac{sin ( x ) - 5 cos ( x ) sin ( x )}{cos ( x )} = 0

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

sin (x) - 5 cos (x) sin (x) = 0

因数

sin (x) - 5 cos (x) sin (x) : - sin (x) (5 cos (x) - 1)

sin (x) - 5 cos (x) sin (x)

共通項をくくり出す

- sin (x)

= - sin (x) (- 1 + 5 cos (x))

- sin (x) (5 cos (x) - 1) = 0

各部分を別個に解く

sin (x) = 0 or 5 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

5 cos (x) - 1 = 0 : x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

5 cos (x) - 1 = 0

1

を右側に移動します

5 cos (x) - 1 = 0

両辺に

1

を足す

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

簡素化

5 cos (x) = 1

5 cos (x) = 1

以下で両辺を割る

5

5 cos (x) = 1

以下で両辺を割る

5

\frac{5 cos ( x )}{5} = \frac{1}{5}

簡素化

cos (x) = \frac{1}{5}

cos (x) = \frac{1}{5}

三角関数の逆数プロパティを適用する

cos (x) = \frac{1}{5}

以下の一般解

cos (x) = \frac{1}{5}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn, x = arccos (\frac{1}{5}) + 2 πn, x = 2 π - arccos (\frac{1}{5}) + 2 πn

10進法形式で解を証明する

x = 2 πn, x = π + 2 πn, x = 1.36943 \dots + 2 πn, x = 2 π - 1.36943 \dots + 2 πn

グラフ

人気の例

cot^2(x)=-2cot(x)-1

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

cos (\frac{π t}{2}) = 0

sin (θ) = - \frac{3}{7}

tan(θ)=(sqrt(3))/(-1)

tan (θ) = \frac{3}{- 1}

cos (x) = \frac{8}{10}