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 >

solvefor x,r^2=9cos(2x)

解

解く

x, r^{2} = 9 cos (2 x)

解

x = \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn, x = - \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

解答ステップ

r^{2} = 9 cos (2 x)

辺を交換する

9 cos (2 x) = r^{2}

以下で両辺を割る

9

9 cos (2 x) = r^{2}

以下で両辺を割る

9

\frac{9 cos ( 2 x )}{9} = \frac{r ^{2}}{9}

簡素化

cos (2 x) = \frac{r ^{2}}{9}

cos (2 x) = \frac{r ^{2}}{9}

三角関数の逆数プロパティを適用する

cos (2 x) = \frac{r ^{2}}{9}

以下の一般解

cos (2 x) = \frac{r ^{2}}{9}

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

2 x = arccos (\frac{r ^{2}}{9}) + 2 πn, 2 x = - arccos (\frac{r ^{2}}{9}) + 2 πn

2 x = arccos (\frac{r ^{2}}{9}) + 2 πn, 2 x = - arccos (\frac{r ^{2}}{9}) + 2 πn

解く

2 x = arccos (\frac{r ^{2}}{9}) + 2 πn : x = \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

2 x = arccos (\frac{r ^{2}}{9}) + 2 πn

以下で両辺を割る

2

2 x = arccos (\frac{r ^{2}}{9}) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = \frac{arccos ( \frac{r ^{2}}{9} )}{2} + \frac{2 πn}{2}

簡素化

x = \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

x = \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

解く

2 x = - arccos (\frac{r ^{2}}{9}) + 2 πn : x = - \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

2 x = - arccos (\frac{r ^{2}}{9}) + 2 πn

以下で両辺を割る

2

2 x = - arccos (\frac{r ^{2}}{9}) + 2 πn

以下で両辺を割る

2

\frac{2 x}{2} = - \frac{arccos ( \frac{r ^{2}}{9} )}{2} + \frac{2 πn}{2}

簡素化

x = - \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

x = - \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

x = \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn, x = - \frac{arccos ( \frac{r ^{2}}{9} )}{2} + πn

グラフ

人気の例

csc (a) = \frac{13}{12}

(4sin(x)-3cos(x))/(3sin(x)+4cos(x))=3

\frac{4 sin ( x ) - 3 cos ( x )}{3 sin ( x ) + 4 cos ( x )} = 3

7=10sin(40(0-h))+17

7 = 10 sin (40 (0 - h)) + 17

sec (x) = - \frac{5}{3}

2sin(2x)+2cos(x)=0

2 sin (2 x) + 2 cos (x) = 0