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人気のある三角関数 >

tan^2(θ)=-3/2 sec(θ),0<= θ<2pi

解

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

解

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

+1

度

θ = 12 0^{\circ}, θ = 24 0^{\circ}

解答ステップ

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

両辺から

- \frac{3}{2} sec (θ)

を引く

tan^{2} (θ) + \frac{3}{2} sec (θ) = 0

簡素化

tan^{2} (θ) + \frac{3}{2} sec (θ) : \frac{2 tan ^{2} ( θ ) + 3 sec ( θ )}{2}

tan^{2} (θ) + \frac{3}{2} sec (θ)

乗じる

\frac{3}{2} sec (θ) : \frac{3 sec ( θ )}{2}

\frac{3}{2} sec (θ)

分数を乗じる:

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

= \frac{3 sec ( θ )}{2}

= tan^{2} (θ) + \frac{3 sec ( θ )}{2}

元を分数に変換する:

tan^{2} (θ) = \frac{tan ^{2} ( θ ) 2}{2}

= \frac{tan ^{2} ( θ ) \cdot 2}{2} + \frac{3 sec ( θ )}{2}

分母が等しいので, 分数を組み合わせる:

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

= \frac{tan ^{2} ( θ ) \cdot 2 + 3 sec ( θ )}{2}

\frac{2 tan ^{2} ( θ ) + 3 sec ( θ )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 tan^{2} (θ) + 3 sec (θ) = 0

三角関数の公式を使用して書き換える

2 tan^{2} (θ) + 3 sec (θ)

ピタゴラスの公式を使用する:

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

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

= 2 (sec^{2} (θ) - 1) + 3 sec (θ)

(- 1 + sec^{2} (θ)) \cdot 2 + 3 sec (θ) = 0

置換で解く

(- 1 + sec^{2} (θ)) \cdot 2 + 3 sec (θ) = 0

仮定:

sec (θ) = u

(- 1 + u^{2}) \cdot 2 + 3 u = 0

(- 1 + u^{2}) \cdot 2 + 3 u = 0 : u = \frac{1}{2}, u = - 2

(- 1 + u^{2}) \cdot 2 + 3 u = 0

拡張

(- 1 + u^{2}) \cdot 2 + 3 u : - 2 + 2 u^{2} + 3 u

(- 1 + u^{2}) \cdot 2 + 3 u

= 2 (- 1 + u^{2}) + 3 u

拡張

2 (- 1 + u^{2}) : - 2 + 2 u^{2}

2 (- 1 + u^{2})

分配法則を適用する:

a (b + c) = ab + a c

a = 2, b = - 1, c = u^{2}

= 2 (- 1) + 2 u^{2}

マイナス・プラスの規則を適用する

+ (- a) = - a

= - 2 \cdot 1 + 2 u^{2}

数を乗じる:

2 \cdot 1 = 2

= - 2 + 2 u^{2}

= - 2 + 2 u^{2} + 3 u

- 2 + 2 u^{2} + 3 u = 0

標準的な形式で書く

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

2 u^{2} + 3 u - 2 = 0

解くとthe二次式

2 u^{2} + 3 u - 2 = 0

二次Equationの公式:

次の場合:

a = 2, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 2) = 5

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

規則を適用

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

数を乗じる:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

数を足す:

9 + 16 = 25

= 25

数を因数に分解する:

25 = 5^{2}

= 5^{2}

累乗根の規則を適用する:

n a^{n} = a

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

解を分離する

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

数を足す/引く:

- 3 + 5 = 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{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

数を引く:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 8}{4}

分数の規則を適用する:

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

= - \frac{8}{4}

数を割る:

\frac{8}{4} = 2

= - 2

二次equationの解:

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

代用を戻す

u = sec (θ)

sec (θ) = \frac{1}{2}, sec (θ) = - 2

sec (θ) = \frac{1}{2}, sec (θ) = - 2

sec (θ) = \frac{1}{2}, 0 \leq θ < 2 π :

解なし

sec (θ) = \frac{1}{2}, 0 \leq θ < 2 π

sec (x) \leq - 1 or sec (x) \geq 1

解なし

sec (θ) = - 2, 0 \leq θ < 2 π : θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

sec (θ) = - 2, 0 \leq θ < 2 π

以下の一般解

sec (θ) = - 2

sec (x)

2 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

範囲の解答

0 \leq θ < 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

すべての解を組み合わせる

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

グラフ

人気の例

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