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人気のある三角関数 >

3tan^4(x)-10tan^2(x)+3=0

解

3 tan^{4} (x) - 10 tan^{2} (x) + 3 = 0

解

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

+1

度

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

解答ステップ

3 tan^{4} (x) - 10 tan^{2} (x) + 3 = 0

置換で解く

3 tan^{4} (x) - 10 tan^{2} (x) + 3 = 0

仮定:

tan (x) = u

3 u^{4} - 10 u^{2} + 3 = 0

3 u^{4} - 10 u^{2} + 3 = 0 : u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{4} - 10 u^{2} + 3 = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

3 v^{2} - 10 v + 3 = 0

解く

3 v^{2} - 10 v + 3 = 0 : v = 3, v = \frac{1}{3}

3 v^{2} - 10 v + 3 = 0

解くとthe二次式

3 v^{2} - 10 v + 3 = 0

二次Equationの公式:

次の場合:

a = 3, b = - 10, c = 3

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

v_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 \cdot 3 = 8

(- 10)^{2} - 4 \cdot 3 \cdot 3

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

(- 10)^{2} = 1 0^{2}

= 1 0^{2} - 4 \cdot 3 \cdot 3

数を乗じる:

4 \cdot 3 \cdot 3 = 36

= 1 0^{2} - 36

1 0^{2} = 100

= 100 - 36

数を引く:

100 - 36 = 64

= 64

数を因数に分解する:

64 = 8^{2}

= 8^{2}

累乗根の規則を適用する:

n a^{n} = a

8^{2} = 8

= 8

v_{1, 2} = \frac{- ( - 10 ) \pm 8}{2 \cdot 3}

解を分離する

v_{1} = \frac{- ( - 10 ) + 8}{2 \cdot 3}, v_{2} = \frac{- ( - 10 ) - 8}{2 \cdot 3}

v = \frac{- ( - 10 ) + 8}{2 \cdot 3} : 3

\frac{- ( - 10 ) + 8}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{10 + 8}{2 \cdot 3}

数を足す:

10 + 8 = 18

= \frac{18}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{18}{6}

数を割る:

\frac{18}{6} = 3

= 3

v = \frac{- ( - 10 ) - 8}{2 \cdot 3} : \frac{1}{3}

\frac{- ( - 10 ) - 8}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{10 - 8}{2 \cdot 3}

数を引く:

10 - 8 = 2

= \frac{2}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{2}{6}

共通因数を約分する:

2

= \frac{1}{3}

二次equationの解:

v = 3, v = \frac{1}{3}

v = 3, v = \frac{1}{3}

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = 3 : u = 3, u = - 3

u^{2} = 3

x^{2} = f (a)

の場合, 解は

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

u = 3, u = - 3

解く

u^{2} = \frac{1}{3} : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} = \frac{1}{3}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{3}, u = - \frac{1}{3}

解答は

u = 3, u = - 3, u = \frac{1}{3}, u = - \frac{1}{3}

代用を戻す

u = tan (x)

tan (x) = 3, tan (x) = - 3, tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = 3, tan (x) = - 3, tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

以下の一般解

tan (x) = 3

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{π}{3} + πn

x = \frac{π}{3} + πn

tan (x) = - 3 : x = \frac{2 π}{3} + πn

tan (x) = - 3

以下の一般解

tan (x) = - 3

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{2 π}{3} + πn

x = \frac{2 π}{3} + πn

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

tan (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

tan (x) = \frac{1}{3}

以下の一般解

tan (x) = \frac{1}{3}

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

三角関数の逆数プロパティを適用する

tan (x) = - \frac{1}{3}

以下の一般解

tan (x) = - \frac{1}{3}

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

すべての解を組み合わせる

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

10進法形式で解を証明する

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

グラフ

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