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人気のある >

tan(x)= 1/(sin(2x)),0<= x<= pi

解

tan (x) = \frac{1}{sin ( 2 x )}, 0 \leq x \leq π

解

x = \frac{π}{4}, x = \frac{3 π}{4}

+1

度

x = 4 5^{\circ}, x = 13 5^{\circ}

解答ステップ

tan (x) = \frac{1}{sin ( 2 x )}, 0 \leq x \leq π

両辺から

\frac{1}{sin ( 2 x )}

を引く

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

簡素化

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

tan (x) - \frac{1}{sin ( 2 x )}

元を分数に変換する:

tan (x) = \frac{tan ( x ) sin ( 2 x )}{sin ( 2 x )}

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

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

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

tan (x) sin (2 x) - 1 = 0

三角関数の公式を使用して書き換える

- 1 + sin (2 x) tan (x)

2倍角の公式を使用:

sin (2 x) = 2 sin (x) cos (x)

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

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

2 cos (x) sin (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 ) \cdot 2 cos ( x ) sin ( x )}{cos ( x )}

共通因数を約分する:

cos (x)

= sin (x) \cdot 2 sin (x)

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

= 2 sin^{2} (x)

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

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

置換で解く

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

仮定:

sin (x) = u

- 1 + 2 u^{2} = 0

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

- 1 + 2 u^{2} = 0

1

を右側に移動します

- 1 + 2 u^{2} = 0

両辺に

1

を足す

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

簡素化

2 u^{2} = 1

2 u^{2} = 1

以下で両辺を割る

2

2 u^{2} = 1

以下で両辺を割る

2

\frac{2 u ^{2}}{2} = \frac{1}{2}

簡素化

u^{2} = \frac{1}{2}

u^{2} = \frac{1}{2}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{2}, u = - \frac{1}{2}

代用を戻す

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, 0 \leq x \leq π : x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = \frac{1}{2}, 0 \leq x \leq π

以下の一般解

sin (x) = \frac{1}{2}

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 = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

範囲の解答

0 \leq x \leq π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq π :

解なし

sin (x) = - \frac{1}{2}, 0 \leq x \leq π

以下の一般解

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

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 = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

範囲の解答

0 \leq x \leq π

解なし

すべての解を組み合わせる

x = \frac{π}{4}, x = \frac{3 π}{4}

グラフ

人気の例

cos (x) = 9054

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

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

6sin^2(θ/2)=cos^2(θ)+5

6 sin^{2} (\frac{θ}{2}) = cos^{2} (θ) + 5

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

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

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