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人気のある >

50=25sin(pi/2 (x+1))+40

解

50 = 25 sin (\frac{π}{2} (x + 1)) + 40

解

x = \frac{2 \cdot 0.41151 \dots}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 \cdot 0.41151 \dots}{π}

+1

度

x = - 42.28544 \dots^{\circ} + 229.18311 \dots^{\circ} n, x = 42.28544 \dots^{\circ} + 229.18311 \dots^{\circ} n

解答ステップ

50 = 25 sin (\frac{π}{2} (x + 1)) + 40

辺を交換する

25 sin (\frac{π}{2} (x + 1)) + 40 = 50

40

を右側に移動します

25 sin (\frac{π}{2} (x + 1)) + 40 = 50

両辺から

40

を引く

25 sin (\frac{π}{2} (x + 1)) + 40 - 40 = 50 - 40

簡素化

25 sin (\frac{π}{2} (x + 1)) = 10

25 sin (\frac{π}{2} (x + 1)) = 10

以下で両辺を割る

25

25 sin (\frac{π}{2} (x + 1)) = 10

以下で両辺を割る

25

\frac{25 sin ( \frac{π}{2} ( x + 1 ) )}{25} = \frac{10}{25}

簡素化

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

三角関数の逆数プロパティを適用する

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

以下の一般解

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

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

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

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

解く

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn : x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

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

以下で両辺を乗じる:

2

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

以下で両辺を乗じる:

2

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

簡素化

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

簡素化

2 \cdot \frac{π}{2} (x + 1) : π (x + 1)

2 \cdot \frac{π}{2} (x + 1)

分数を乗じる:

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

= \frac{2 π}{2} (x + 1)

共通因数を約分する:

2

= (x + 1) π

簡素化

2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn : 2 arcsin (\frac{2}{5}) + 4 πn

2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

以下で両辺を割る

π

π (x + 1) = 2 arcsin (\frac{2}{5}) + 4 πn

以下で両辺を割る

π

\frac{π ( x + 1 )}{π} = \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

簡素化

x + 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

1

を右側に移動します

x + 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

両辺から

1

を引く

x + 1 - 1 = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

簡素化

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

解く

\frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn : x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

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

以下で両辺を乗じる:

2

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

以下で両辺を乗じる:

2

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

簡素化

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

簡素化

2 \cdot \frac{π}{2} (x + 1) : π (x + 1)

2 \cdot \frac{π}{2} (x + 1)

分数を乗じる:

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

= \frac{2 π}{2} (x + 1)

共通因数を約分する:

2

= (x + 1) π

簡素化

2 π - 2 arcsin (\frac{2}{5}) + 2 \cdot 2 πn : 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

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

数を乗じる:

2 \cdot 2 = 4

= 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

以下で両辺を割る

π

π (x + 1) = 2 π - 2 arcsin (\frac{2}{5}) + 4 πn

以下で両辺を割る

π

\frac{π ( x + 1 )}{π} = \frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

簡素化

\frac{π ( x + 1 )}{π} = \frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

簡素化

\frac{π ( x + 1 )}{π} : x + 1

\frac{π ( x + 1 )}{π}

共通因数を約分する:

π

= x + 1

簡素化

\frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π} : 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

\frac{2 π}{π} - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

キャンセル

\frac{2 π}{π} : 2

\frac{2 π}{π}

共通因数を約分する:

π

= 2

= 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + \frac{4 πn}{π}

キャンセル

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

共通因数を約分する:

π

= 4 n

= 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

1

を右側に移動します

x + 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n

両辺から

1

を引く

x + 1 - 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

簡素化

x + 1 - 1 = 2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

簡素化

x + 1 - 1 : x

x + 1 - 1

類似した元を足す:

1 - 1 = 0

= x

簡素化

2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1 : 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

2 - \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

数を引く:

2 - 1 = 1

= 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

10進法形式で解を証明する

x = \frac{2 \cdot 0.41151 \dots}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 \cdot 0.41151 \dots}{π}

グラフ

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