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人気のある三角関数 >

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

解

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

解

x = 1.10714 \dots + πn, x = \frac{π}{4} + πn

+1

度

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

tan (x) = u

u (u - 3) + 2 = 0

u (u - 3) + 2 = 0 : u = 2, u = 1

u (u - 3) + 2 = 0

拡張

u (u - 3) + 2 : u^{2} - 3 u + 2

u (u - 3) + 2

拡張

u (u - 3) : u^{2} - 3 u

u (u - 3)

分配法則を適用する:

a (b - c) = ab - a c

a = u, b = u, c = 3

= uu - u \cdot 3

= uu - 3 u

uu = u^{2}

uu

指数の規則を適用する:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

数を足す:

1 + 1 = 2

= u^{2}

= u^{2} - 3 u

= u^{2} - 3 u + 2

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 1, b = - 3, c = 2

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

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

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

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

指数の規則を適用する:

n

が偶数であれば

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

(- 3)^{2} = 3^{2}

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

数を乗じる:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数を引く:

9 - 8 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

3 + 1 = 4

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

数を乗じる:

2 \cdot 1 = 2

= \frac{4}{2}

数を割る:

\frac{4}{2} = 2

= 2

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

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

規則を適用

- (- a) = a

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

数を引く:

3 - 1 = 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:

u = 2, u = 1

代用を戻す

u = tan (x)

tan (x) = 2, tan (x) = 1

tan (x) = 2, tan (x) = 1

tan (x) = 2 : x = arctan (2) + πn

tan (x) = 2

三角関数の逆数プロパティを適用する

tan (x) = 2

以下の一般解

tan (x) = 2

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

x = arctan (2) + πn

x = arctan (2) + 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

すべての解を組み合わせる

x = arctan (2) + πn, x = \frac{π}{4} + πn

10進法形式で解を証明する

x = 1.10714 \dots + πn, x = \frac{π}{4} + πn

グラフ

人気の例

sin(x)-4sin^3(x)=0

sin (x) - 4 sin^{3} (x) = 0

2sin(x)=3sqrt(3)

2 sin (x) = 33

sin^{2} (θ) = \frac{1}{9}

(sec(B)+csc(B))/(1+tan(B))=csc^2(B)

\frac{sec ( B ) + csc ( B )}{1 + tan ( B )} = csc^{2} (B)

4=sqrt(2)csc(3x)

4 = 2 csc (3 x)