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人気のある三角関数 >

(cos^2(a)-3cos(a)+2)/(sin^2(a))=1

解

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

解

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

+1

度

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n

解答ステップ

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} = 1

両辺から

1

を引く

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 = 0

簡素化

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1 : \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )}

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - 1

元を分数に変換する:

1 = \frac{1 sin ^{2} ( a )}{sin ^{2} ( a )}

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2}{sin ^{2} ( a )} - \frac{1 \cdot sin ^{2} ( a )}{sin ^{2} ( a )}

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{cos ^{2} ( a ) - 3 cos ( a ) + 2 - 1 \cdot sin ^{2} ( a )}{sin ^{2} ( a )}

乗算:

1 \cdot sin^{2} (a) = sin^{2} (a)

= \frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )}

\frac{cos ^{2} ( a ) - 3 cos ( a ) + 2 - sin ^{2} ( a )}{sin ^{2} ( a )} = 0

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

cos^{2} (a) - 3 cos (a) + 2 - sin^{2} (a) = 0

三角関数の公式を使用して書き換える

2 + cos^{2} (a) - sin^{2} (a) - 3 cos (a)

ピタゴラスの公式を使用する:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a)

簡素化

2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a) : 2 cos^{2} (a) - 3 cos (a) + 1

2 + cos^{2} (a) - (1 - cos^{2} (a)) - 3 cos (a)

- (1 - cos^{2} (a)) : - 1 + cos^{2} (a)

- (1 - cos^{2} (a))

括弧を分配する

= - (1) - (- cos^{2} (a))

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (a)

= 2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a)

簡素化

2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a) : 2 cos^{2} (a) - 3 cos (a) + 1

2 + cos^{2} (a) - 1 + cos^{2} (a) - 3 cos (a)

条件のようなグループ

= cos^{2} (a) + cos^{2} (a) - 3 cos (a) + 2 - 1

類似した元を足す:

cos^{2} (a) + cos^{2} (a) = 2 cos^{2} (a)

= 2 cos^{2} (a) - 3 cos (a) + 2 - 1

数を足す/引く:

2 - 1 = 1

= 2 cos^{2} (a) - 3 cos (a) + 1

= 2 cos^{2} (a) - 3 cos (a) + 1

= 2 cos^{2} (a) - 3 cos (a) + 1

1 + 2 cos^{2} (a) - 3 cos (a) = 0

置換で解く

1 + 2 cos^{2} (a) - 3 cos (a) = 0

仮定:

cos (a) = u

1 + 2 u^{2} - 3 u = 0

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

1 + 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 + 1 = 0

解くとthe二次式

2 u^{2} - 3 u + 1 = 0

二次Equationの公式:

次の場合:

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

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

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

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

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

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 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 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 2}

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

3 + 1 = 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

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

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

規則を適用

- (- a) = a

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

数を引く:

3 - 1 = 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}

二次equationの解:

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

代用を戻す

u = cos (a)

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1, cos (a) = \frac{1}{2}

cos (a) = 1 : a = 2 πn

cos (a) = 1

以下の一般解

cos (a) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = 0 + 2 πn

a = 0 + 2 πn

解く

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 2 πn

cos (a) = \frac{1}{2} : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

cos (a) = \frac{1}{2}

以下の一般解

cos (a) = \frac{1}{2}

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

すべての解を組み合わせる

a = 2 πn, a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

equationは以下で未定義のため:

2 πn

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

グラフ

人気の例

(sin(x)-(sqrt(2)))/2 =0

\frac{sin ( x ) - ( 2 )}{2} = 0

cos(2x)=5-6cos^2(x)

cos (2 x) = 5 - 6 cos^{2} (x)

cos^{4} (x) = 0.37

sin^2(x)=((sqrt(3)))/2

sin^{2} (x) = \frac{( 3 )}{2}

sin(x)cos(x)=2sin(2x)

sin (x) cos (x) = 2 sin (2 x)