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人気のある三角関数 >

sin(2t-pi/5)=-1/3

解

sin (2 t - \frac{π}{5}) = - \frac{1}{3}

解

t = πn + \frac{π}{10} - \frac{0.33983 \dots}{2}, t = πn + \frac{π}{2} + \frac{π}{10} + \frac{0.33983 \dots}{2}

+1

度

t = 8.26438 \dots^{\circ} + 18 0^{\circ} n, t = 117.73561 \dots^{\circ} + 18 0^{\circ} n

解答ステップ

sin (2 t - \frac{π}{5}) = - \frac{1}{3}

三角関数の逆数プロパティを適用する

sin (2 t - \frac{π}{5}) = - \frac{1}{3}

以下の一般解

sin (2 t - \frac{π}{5}) = - \frac{1}{3}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

2 t - \frac{π}{5} = arcsin (- \frac{1}{3}) + 2 πn, 2 t - \frac{π}{5} = π + arcsin (\frac{1}{3}) + 2 πn

2 t - \frac{π}{5} = arcsin (- \frac{1}{3}) + 2 πn, 2 t - \frac{π}{5} = π + arcsin (\frac{1}{3}) + 2 πn

解く

2 t - \frac{π}{5} = arcsin (- \frac{1}{3}) + 2 πn : t = πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}

2 t - \frac{π}{5} = arcsin (- \frac{1}{3}) + 2 πn

簡素化

arcsin (- \frac{1}{3}) + 2 πn : - arcsin (\frac{1}{3}) + 2 πn

arcsin (- \frac{1}{3}) + 2 πn

次のプロパティを使用する:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{3}) = - arcsin (\frac{1}{3})

= - arcsin (\frac{1}{3}) + 2 πn

2 t - \frac{π}{5} = - arcsin (\frac{1}{3}) + 2 πn

\frac{π}{5}

を右側に移動します

2 t - \frac{π}{5} = - arcsin (\frac{1}{3}) + 2 πn

両辺に

\frac{π}{5}

を足す

2 t - \frac{π}{5} + \frac{π}{5} = - arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

簡素化

2 t = - arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

2 t = - arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

以下で両辺を割る

2

2 t = - arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

以下で両辺を割る

2

\frac{2 t}{2} = - \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2}

簡素化

\frac{2 t}{2} = - \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2}

簡素化

\frac{2 t}{2} : t

\frac{2 t}{2}

数を割る:

\frac{2}{2} = 1

= t

簡素化

- \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2} : πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}

- \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2}

条件のようなグループ

= \frac{2 πn}{2} + \frac{\frac{π}{5}}{2} - \frac{arcsin ( \frac{1}{3} )}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{5}}{2} = \frac{π}{10}

\frac{\frac{π}{5}}{2}

分数の規則を適用する:

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

= \frac{π}{5 \cdot 2}

数を乗じる:

5 \cdot 2 = 10

= \frac{π}{10}

= πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}

解く

2 t - \frac{π}{5} = π + arcsin (\frac{1}{3}) + 2 πn : t = πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

2 t - \frac{π}{5} = π + arcsin (\frac{1}{3}) + 2 πn

\frac{π}{5}

を右側に移動します

2 t - \frac{π}{5} = π + arcsin (\frac{1}{3}) + 2 πn

両辺に

\frac{π}{5}

を足す

2 t - \frac{π}{5} + \frac{π}{5} = π + arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

簡素化

2 t = π + arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

2 t = π + arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

以下で両辺を割る

2

2 t = π + arcsin (\frac{1}{3}) + 2 πn + \frac{π}{5}

以下で両辺を割る

2

\frac{2 t}{2} = \frac{π}{2} + \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2}

簡素化

\frac{2 t}{2} = \frac{π}{2} + \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2}

簡素化

\frac{2 t}{2} : t

\frac{2 t}{2}

数を割る:

\frac{2}{2} = 1

= t

簡素化

\frac{π}{2} + \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2} : πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

\frac{π}{2} + \frac{arcsin ( \frac{1}{3} )}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2}

条件のようなグループ

= \frac{π}{2} + \frac{2 πn}{2} + \frac{\frac{π}{5}}{2} + \frac{arcsin ( \frac{1}{3} )}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数を割る:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{5}}{2} = \frac{π}{10}

\frac{\frac{π}{5}}{2}

分数の規則を適用する:

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

= \frac{π}{5 \cdot 2}

数を乗じる:

5 \cdot 2 = 10

= \frac{π}{10}

= \frac{π}{2} + πn + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

条件のようなグループ

= πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

t = πn + \frac{π}{10} - \frac{arcsin ( \frac{1}{3} )}{2}, t = πn + \frac{π}{2} + \frac{π}{10} + \frac{arcsin ( \frac{1}{3} )}{2}

10進法形式で解を証明する

t = πn + \frac{π}{10} - \frac{0.33983 \dots}{2}, t = πn + \frac{π}{2} + \frac{π}{10} + \frac{0.33983 \dots}{2}

グラフ

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