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人気のある三角関数 >

(cos^4(x))/3 =sin^2(x)

解

\frac{cos ^{4} ( x )}{3} = sin^{2} (x)

解

x = - 0.47445 \dots + 2 πn, x = π + 0.47445 \dots + 2 πn, x = 0.47445 \dots + 2 πn, x = π - 0.47445 \dots + 2 πn

+1

度

x = - 27.18404 \dots^{\circ} + 36 0^{\circ} n, x = 207.18404 \dots^{\circ} + 36 0^{\circ} n, x = 27.18404 \dots^{\circ} + 36 0^{\circ} n, x = 152.81595 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

\frac{cos ^{4} ( x )}{3} = sin^{2} (x)

両辺から

sin^{2} (x)

を引く

\frac{cos ^{4} ( x )}{3} - sin^{2} (x) = 0

簡素化

\frac{cos ^{4} ( x )}{3} - sin^{2} (x) : \frac{cos ^{4} ( x ) - 3 sin ^{2} ( x )}{3}

\frac{cos ^{4} ( x )}{3} - sin^{2} (x)

元を分数に変換する:

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

= \frac{cos ^{4} ( x )}{3} - \frac{sin ^{2} ( x ) \cdot 3}{3}

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 ^{4} ( x ) - sin ^{2} ( x ) \cdot 3}{3}

\frac{cos ^{4} ( x ) - 3 sin ^{2} ( x )}{3} = 0

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

cos^{4} (x) - 3 sin^{2} (x) = 0

因数

cos^{4} (x) - 3 sin^{2} (x) : (cos^{2} (x) + 3 sin (x)) (cos^{2} (x) - 3 sin (x))

cos^{4} (x) - 3 sin^{2} (x)

cos^{4} (x) - 3 sin^{2} (x)

を書き換え

(cos^{2} (x))^{2} - (3 sin (x))^{2}

cos^{4} (x) - 3 sin^{2} (x)

累乗根の規則を適用する:

a = (a)^{2}

3 = (3)^{2}

= cos^{4} (x) - (3)^{2} sin^{2} (x)

指数の規則を適用する:

a^{b c} = (a^{b})^{c}

cos^{4} (x) = (cos^{2} (x))^{2}

= (cos^{2} (x))^{2} - (3)^{2} sin^{2} (x)

指数の規則を適用する:

a^{m} b^{m} = (ab)^{m}

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

= (cos^{2} (x))^{2} - (3 sin (x))^{2}

= (cos^{2} (x))^{2} - (3 sin (x))^{2}

2乗の差の公式を適用する:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

各部分を別個に解く

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

cos^{2} (x) + 3 sin (x) = 0 : x = arcsin (- \frac{- 3 + 7}{2}) + 2 πn, x = π + arcsin (\frac{- 3 + 7}{2}) + 2 πn

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

三角関数の公式を使用して書き換える

cos^{2} (x) + sin (x) 3

ピタゴラスの公式を使用する:

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

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

= 1 - sin^{2} (x) + sin (x) 3

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

置換で解く

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

仮定:

sin (x) = u

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

1 - u^{2} + u 3 = 0 : u = - \frac{- 3 + 7}{2}, u = \frac{3 + 7}{2}

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

標準的な形式で書く

a x^{2} + b x + c = 0

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

(3)^{2} - 4 (- 1) \cdot 1 = 7

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

規則を適用

- (- a) = a

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

(3)^{2} = 3

(3)^{2}

累乗根の規則を適用する:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

共通因数を約分する:

2

= 1

= 3

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 4

= 3 + 4

数を足す:

3 + 4 = 7

= 7

u_{1, 2} = \frac{- 3 \pm 7}{2 ( - 1 )}

解を分離する

u_{1} = \frac{- 3 + 7}{2 ( - 1 )}, u_{2} = \frac{- 3 - 7}{2 ( - 1 )}

u = \frac{- 3 + 7}{2 ( - 1 )} : - \frac{- 3 + 7}{2}

\frac{- 3 + 7}{2 ( - 1 )}

括弧を削除する:

(- a) = - a

= \frac{- 3 + 7}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 3 + 7}{- 2}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{- 3 + 7}{2}

u = \frac{- 3 - 7}{2 ( - 1 )} : \frac{3 + 7}{2}

\frac{- 3 - 7}{2 ( - 1 )}

括弧を削除する:

(- a) = - a

= \frac{- 3 - 7}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{- 3 - 7}{- 2}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

- 3 - 7 = - (3 + 7)

= \frac{3 + 7}{2}

二次equationの解:

u = - \frac{- 3 + 7}{2}, u = \frac{3 + 7}{2}

代用を戻す

u = sin (x)

sin (x) = - \frac{- 3 + 7}{2}, sin (x) = \frac{3 + 7}{2}

sin (x) = - \frac{- 3 + 7}{2}, sin (x) = \frac{3 + 7}{2}

sin (x) = - \frac{- 3 + 7}{2} : x = arcsin (- \frac{- 3 + 7}{2}) + 2 πn, x = π + arcsin (\frac{- 3 + 7}{2}) + 2 πn

sin (x) = - \frac{- 3 + 7}{2}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{- 3 + 7}{2}

以下の一般解

sin (x) = - \frac{- 3 + 7}{2}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{- 3 + 7}{2}) + 2 πn, x = π + arcsin (\frac{- 3 + 7}{2}) + 2 πn

x = arcsin (- \frac{- 3 + 7}{2}) + 2 πn, x = π + arcsin (\frac{- 3 + 7}{2}) + 2 πn

sin (x) = \frac{3 + 7}{2} :

解なし

sin (x) = \frac{3 + 7}{2}

- 1 \leq sin (x) \leq 1

解なし

すべての解を組み合わせる

x = arcsin (- \frac{- 3 + 7}{2}) + 2 πn, x = π + arcsin (\frac{- 3 + 7}{2}) + 2 πn

cos^{2} (x) - 3 sin (x) = 0 : x = arcsin (\frac{7 - 3}{2}) + 2 πn, x = π - arcsin (\frac{7 - 3}{2}) + 2 πn

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

三角関数の公式を使用して書き換える

cos^{2} (x) - sin (x) 3

ピタゴラスの公式を使用する:

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

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

= 1 - sin^{2} (x) - sin (x) 3

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

置換で解く

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

仮定:

sin (x) = u

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

1 - u^{2} - u 3 = 0 : u = - \frac{3 + 7}{2}, u = \frac{7 - 3}{2}

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

標準的な形式で書く

a x^{2} + b x + c = 0

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

(- 3)^{2} - 4 (- 1) \cdot 1 = 7

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

規則を適用

- (- a) = a

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

(- 3)^{2} = 3

(- 3)^{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}

累乗根の規則を適用する:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

共通因数を約分する:

2

= 1

= 3

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数を乗じる:

4 \cdot 1 \cdot 1 = 4

= 4

= 3 + 4

数を足す:

3 + 4 = 7

= 7

u_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 ( - 1 )}

解を分離する

u_{1} = \frac{- ( - 3 ) + 7}{2 ( - 1 )}, u_{2} = \frac{- ( - 3 ) - 7}{2 ( - 1 )}

u = \frac{- ( - 3 ) + 7}{2 ( - 1 )} : - \frac{3 + 7}{2}

\frac{- ( - 3 ) + 7}{2 ( - 1 )}

括弧を削除する:

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

= \frac{3 + 7}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{3 + 7}{- 2}

分数の規則を適用する:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{3 + 7}{2}

u = \frac{- ( - 3 ) - 7}{2 ( - 1 )} : \frac{7 - 3}{2}

\frac{- ( - 3 ) - 7}{2 ( - 1 )}

括弧を削除する:

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

= \frac{3 - 7}{- 2 \cdot 1}

数を乗じる:

2 \cdot 1 = 2

= \frac{3 - 7}{- 2}

分数の規則を適用する:

\frac{- a}{- b} = \frac{a}{b}

3 - 7 = - (7 - 3)

= \frac{7 - 3}{2}

二次equationの解:

u = - \frac{3 + 7}{2}, u = \frac{7 - 3}{2}

代用を戻す

u = sin (x)

sin (x) = - \frac{3 + 7}{2}, sin (x) = \frac{7 - 3}{2}

sin (x) = - \frac{3 + 7}{2}, sin (x) = \frac{7 - 3}{2}

sin (x) = - \frac{3 + 7}{2} :

解なし

sin (x) = - \frac{3 + 7}{2}

- 1 \leq sin (x) \leq 1

解なし

sin (x) = \frac{7 - 3}{2} : x = arcsin (\frac{7 - 3}{2}) + 2 πn, x = π - arcsin (\frac{7 - 3}{2}) + 2 πn

sin (x) = \frac{7 - 3}{2}

三角関数の逆数プロパティを適用する

sin (x) = \frac{7 - 3}{2}

以下の一般解

sin (x) = \frac{7 - 3}{2}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{7 - 3}{2}) + 2 πn, x = π - arcsin (\frac{7 - 3}{2}) + 2 πn

x = arcsin (\frac{7 - 3}{2}) + 2 πn, x = π - arcsin (\frac{7 - 3}{2}) + 2 πn

すべての解を組み合わせる

x = arcsin (\frac{7 - 3}{2}) + 2 πn, x = π - arcsin (\frac{7 - 3}{2}) + 2 πn

すべての解を組み合わせる

x = arcsin (- \frac{- 3 + 7}{2}) + 2 πn, x = π + arcsin (\frac{- 3 + 7}{2}) + 2 πn, x = arcsin (\frac{7 - 3}{2}) + 2 πn, x = π - arcsin (\frac{7 - 3}{2}) + 2 πn

10進法形式で解を証明する

x = - 0.47445 \dots + 2 πn, x = π + 0.47445 \dots + 2 πn, x = 0.47445 \dots + 2 πn, x = π - 0.47445 \dots + 2 πn

グラフ

人気の例

sin^3(x)cos(x)-sin^2(x)=0

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

sin^2(x)+sin^4(x)=0

sin^{2} (x) + sin^{4} (x) = 0

cos^2(45-a)-sin^2(45-a)=sin^2(a)

cos^{2} (4 5^{\circ} - a) - sin^{2} (4 5^{\circ} - a) = sin^{2} (a)

sin^2(2x)-cos^2(2x)=0

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

cos^2(x)cos(x)=0

cos^{2} (x) cos (x) = 0