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人気のある三角関数 >

2sec^2(x)=3-tan(x),0<= x<= 2pi

解

2 sec^{2} (x) = 3 - tan (x), 0 \leq x \leq 2 π

解

x = 0.46364 \dots, x = 0.46364 \dots + π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

+1

度

x = 26.56505 \dots^{\circ}, x = 206.56505 \dots^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

解答ステップ

2 sec^{2} (x) = 3 - tan (x), 0 \leq x \leq 2 π

両辺から

3 - tan (x)

を引く

2 sec^{2} (x) - 3 + tan (x) = 0

三角関数の公式を使用して書き換える

- 3 + tan (x) + 2 sec^{2} (x)

ピタゴラスの公式を使用する:

sec^{2} (x) = tan^{2} (x) + 1

= - 3 + tan (x) + 2 (tan^{2} (x) + 1)

簡素化

- 3 + tan (x) + 2 (tan^{2} (x) + 1) : 2 tan^{2} (x) + tan (x) - 1

- 3 + tan (x) + 2 (tan^{2} (x) + 1)

拡張

2 (tan^{2} (x) + 1) : 2 tan^{2} (x) + 2

2 (tan^{2} (x) + 1)

分配法則を適用する:

a (b + c) = ab + a c

a = 2, b = tan^{2} (x), c = 1

= 2 tan^{2} (x) + 2 \cdot 1

数を乗じる:

2 \cdot 1 = 2

= 2 tan^{2} (x) + 2

= - 3 + tan (x) + 2 tan^{2} (x) + 2

簡素化

- 3 + tan (x) + 2 tan^{2} (x) + 2 : 2 tan^{2} (x) + tan (x) - 1

- 3 + tan (x) + 2 tan^{2} (x) + 2

条件のようなグループ

= tan (x) + 2 tan^{2} (x) - 3 + 2

数を足す/引く:

- 3 + 2 = - 1

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

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

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

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

置換で解く

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

仮定:

tan (x) = u

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

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

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

標準的な形式で書く

a x^{2} + b x + c = 0

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 2, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 ( - 1 )}{2 \cdot 2}

1^{2} - 4 \cdot 2 (- 1) = 3

1^{2} - 4 \cdot 2 (- 1)

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 (- 1)

規則を適用

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 1 + 3}{2 \cdot 2}, u_{2} = \frac{- 1 - 3}{2 \cdot 2}

u = \frac{- 1 + 3}{2 \cdot 2} : \frac{1}{2}

\frac{- 1 + 3}{2 \cdot 2}

数を足す/引く:

- 1 + 3 = 2

= \frac{2}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 1 - 3}{2 \cdot 2} : - 1

\frac{- 1 - 3}{2 \cdot 2}

数を引く:

- 1 - 3 = - 4

= \frac{- 4}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 4}{4}

分数の規則を適用する:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

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

代用を戻す

u = tan (x)

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

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

tan (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = arctan (\frac{1}{2}), x = arctan (\frac{1}{2}) + π

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{2}) + πn

x = arctan (\frac{1}{2}) + πn

範囲の解答

0 \leq x \leq 2 π

x = arctan (\frac{1}{2}), x = arctan (\frac{1}{2}) + π

tan (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{4}, x = \frac{7 π}{4}

tan (x) = - 1, 0 \leq x \leq 2 π

以下の一般解

tan (x) = - 1

tan (x)

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

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

範囲の解答

0 \leq x \leq 2 π

x = \frac{3 π}{4}, x = \frac{7 π}{4}

すべての解を組み合わせる

x = arctan (\frac{1}{2}), x = arctan (\frac{1}{2}) + π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

10進法形式で解を証明する

x = 0.46364 \dots, x = 0.46364 \dots + π, x = \frac{3 π}{4}, x = \frac{7 π}{4}

グラフ

人気の例

1+cos(2θ)=-3cos(θ)

1 + cos (2 θ) = - 3 cos (θ)

tan (θ) = \frac{6}{0}

3cos(20t)+6sin(20t)=0

3 cos (20 t) + 6 sin (20 t) = 0

tan (θ) = \frac{6}{2}

tan (θ) = \frac{6}{4}