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 >

5-tan^2(x)=4,0<= x<= 360

解

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

解

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

+1

ラジアン

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

解答ステップ

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

置換で解く

5 - tan^{2} (x) = 4

仮定:

tan (x) = u

5 - u^{2} = 4

5 - u^{2} = 4 : u = 1, u = - 1

5 - u^{2} = 4

5

を右側に移動します

5 - u^{2} = 4

両辺から

5

を引く

5 - u^{2} - 5 = 4 - 5

簡素化

- u^{2} = - 1

- u^{2} = - 1

以下で両辺を割る

- 1

- u^{2} = - 1

以下で両辺を割る

- 1

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

簡素化

u^{2} = 1

u^{2} = 1

x^{2} = f (a)

の場合, 解は

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

u = 1, u = - 1

1 = 1

1

規則を適用

1 = 1

= 1

- 1 = - 1

- 1

規則を適用

1 = 1

= - 1

u = 1, u = - 1

代用を戻す

u = tan (x)

tan (x) = 1, tan (x) = - 1

tan (x) = 1, tan (x) = - 1

tan (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 4 5^{\circ}, x = 22 5^{\circ}

tan (x) = 1, 0 \leq x \leq 36 0^{\circ}

以下の一般解

tan (x) = 1

tan (x)

18 0^{\circ} n

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 4 5^{\circ} + 18 0^{\circ} n

x = 4 5^{\circ} + 18 0^{\circ} n

範囲の解答

0 \leq x \leq 36 0^{\circ}

x = 4 5^{\circ}, x = 22 5^{\circ}

tan (x) = - 1, 0 \leq x \leq 36 0^{\circ} : x = 13 5^{\circ}, x = 31 5^{\circ}

tan (x) = - 1, 0 \leq x \leq 36 0^{\circ}

以下の一般解

tan (x) = - 1

tan (x)

18 0^{\circ} n

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

範囲の解答

0 \leq x \leq 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

すべての解を組み合わせる

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

グラフ

人気の例

-2tan(x)+1=2sqrt(3)+1

- 2 tan (x) + 1 = 23 + 1

sin(x-pi/3)= 1/2

sin (x - \frac{π}{3}) = \frac{1}{2}

sec^2(x)-1*sec^2(x)=sin^2(x)

sec^{2} (x) - 1 \cdot sec^{2} (x) = sin^{2} (x)

(sqrt(17))/8 sin(8sqrt(2)x+1.326)=0

\frac{17}{8} sin (82 x + 1.326) = 0

-1-2cos(x)=-cos(x)

- 1 - 2 cos (x) = - cos (x)