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人気のある三角関数 >

(cos^2(x)-1)/(sin^2(x))=-tan(x)

解

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

解

x = \frac{π}{4} + πn

+1

度

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

解答ステップ

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

両辺から

- tan (x)

を引く

\frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} + tan (x) = 0

簡素化

\frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} + tan (x) : \frac{cos ^{2} ( x ) - 1 + sin ^{2} ( x ) tan ( x )}{sin ^{2} ( x )}

\frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} + tan (x)

元を分数に変換する:

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

= \frac{cos ^{2} ( x ) - 1}{sin ^{2} ( x )} + \frac{tan ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

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} ( x ) - 1 + tan ( x ) sin ^{2} ( x )}{sin ^{2} ( x )}

\frac{cos ^{2} ( x ) - 1 + sin ^{2} ( x ) tan ( x )}{sin ^{2} ( x )} = 0

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

cos^{2} (x) - 1 + sin^{2} (x) tan (x) = 0

三角関数の公式を使用して書き換える

- 1 + cos^{2} (x) + sin^{2} (x) tan (x)

ピタゴラスの公式を使用する:

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

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

= sin^{2} (x) tan (x) - sin^{2} (x)

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

因数

- sin^{2} (x) + sin^{2} (x) tan (x) : sin^{2} (x) (tan (x) - 1)

- sin^{2} (x) + sin^{2} (x) tan (x)

共通項をくくり出す

sin^{2} (x)

= sin^{2} (x) (- 1 + tan (x))

sin^{2} (x) (tan (x) - 1) = 0

各部分を別個に解く

sin^{2} (x) = 0 or tan (x) - 1 = 0

sin^{2} (x) = 0 : x = 2 πn, x = π + 2 πn

sin^{2} (x) = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

以下の一般解

sin (x) = 0

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

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

tan (x) - 1 = 0

1

を右側に移動します

tan (x) - 1 = 0

両辺に

1

を足す

tan (x) - 1 + 1 = 0 + 1

簡素化

tan (x) = 1

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

equationは以下で未定義のため:

2 πn, π + 2 πn

x = \frac{π}{4} + πn

グラフ

人気の例

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

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

2sin^3(x)-5sin^2(x)+2sin(x)=0

2 sin^{3} (x) - 5 sin^{2} (x) + 2 sin (x) = 0

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

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

(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)