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人気のある三角関数 >

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

解

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

解

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

+1

度

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

tan (x) = u

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

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

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

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

v^{2} - 4 v + 3 = 0

解く

v^{2} - 4 v + 3 = 0 : v = 3, v = 1

v^{2} - 4 v + 3 = 0

解くとthe二次式

v^{2} - 4 v + 3 = 0

二次Equationの公式:

次の場合:

a = 1, b = - 4, c = 3

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

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

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

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

指数の規則を適用する:

n

が偶数であれば

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

(- 4)^{2} = 4^{2}

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

数を乗じる:

4 \cdot 1 \cdot 3 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数を引く:

16 - 12 = 4

= 4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

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

解を分離する

v_{1} = \frac{- ( - 4 ) + 2}{2 \cdot 1}, v_{2} = \frac{- ( - 4 ) - 2}{2 \cdot 1}

v = \frac{- ( - 4 ) + 2}{2 \cdot 1} : 3

\frac{- ( - 4 ) + 2}{2 \cdot 1}

規則を適用

- (- a) = a

= \frac{4 + 2}{2 \cdot 1}

数を足す:

4 + 2 = 6

= \frac{6}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{6}{2}

数を割る:

\frac{6}{2} = 3

= 3

v = \frac{- ( - 4 ) - 2}{2 \cdot 1} : 1

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

規則を適用

- (- a) = a

= \frac{4 - 2}{2 \cdot 1}

数を引く:

4 - 2 = 2

= \frac{2}{2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{2}{2}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

v = 3, v = 1

v = 3, v = 1

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} = 1 : u = 1, u = - 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

解答は

u = 3, u = - 3, u = 1, u = - 1

代用を戻す

u = tan (x)

tan (x) = 3, tan (x) = - 3, tan (x) = 1, tan (x) = - 1

tan (x) = 3, tan (x) = - 3, tan (x) = 1, tan (x) = - 1

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) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

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{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

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

すべての解を組み合わせる

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

グラフ

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