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人気のある三角関数 >

2(cos^4(x)+1-sin^4(x))=3

解

2 (cos^{4} (x) + 1 - sin^{4} (x)) = 3

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 (cos^{4} (x) + 1 - sin^{4} (x)) = 3

両辺から

3

を引く

2 cos^{4} (x) - 2 sin^{4} (x) - 1 = 0

指数の規則を適用する:

a^{b} = a^{2} a^{b - 2}

- 1 - 2 sin^{4} (x) + 2 cos^{2} (x) cos^{2} (x) = 0

三角関数の公式を使用して書き換える

- 1 - 2 sin^{4} (x) + 2 cos^{2} (x) cos^{2} (x)

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 - 2 sin^{4} (x) + 2 (1 - sin^{2} (x)) (1 - sin^{2} (x))

簡素化

- 1 - 2 sin^{4} (x) + 2 (1 - sin^{2} (x)) (1 - sin^{2} (x)) : - 4 sin^{2} (x) + 1

- 1 - 2 sin^{4} (x) + 2 (1 - sin^{2} (x)) (1 - sin^{2} (x))

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

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

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

= - 1 - 2 sin^{4} (x) + 2 (- sin^{2} (x) + 1)^{2}

(1 - sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

完全平方式を適用する:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = sin^{2} (x)

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

簡素化

1^{2} - 2 \cdot 1 \cdot sin^{2} (x) + (sin^{2} (x))^{2} : 1 - 2 sin^{2} (x) + sin^{4} (x)

1^{2} - 2 \cdot 1 \cdot sin^{2} (x) + (sin^{2} (x))^{2}

規則を適用

1^{a} = 1

1^{2} = 1

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

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

2 \cdot 1 \cdot sin^{2} (x)

数を乗じる:

2 \cdot 1 = 2

= 2 sin^{2} (x)

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

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

指数の規則を適用する:

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

= sin^{2 \cdot 2} (x)

数を乗じる:

2 \cdot 2 = 4

= sin^{4} (x)

= 1 - 2 sin^{2} (x) + sin^{4} (x)

= 1 - 2 sin^{2} (x) + sin^{4} (x)

= - 1 - 2 sin^{4} (x) + 2 (1 - 2 sin^{2} (x) + sin^{4} (x))

拡張

2 (1 - 2 sin^{2} (x) + sin^{4} (x)) : 2 - 4 sin^{2} (x) + 2 sin^{4} (x)

2 (1 - 2 sin^{2} (x) + sin^{4} (x))

括弧を分配する

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

マイナス・プラスの規則を適用する

+ (- a) = - a

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

簡素化

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) + 2 sin^{4} (x) : 2 - 4 sin^{2} (x) + 2 sin^{4} (x)

2 \cdot 1 - 2 \cdot 2 sin^{2} (x) + 2 sin^{4} (x)

数を乗じる:

2 \cdot 1 = 2

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

数を乗じる:

2 \cdot 2 = 4

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

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

= - 1 - 2 sin^{4} (x) + 2 - 4 sin^{2} (x) + 2 sin^{4} (x)

簡素化

- 1 - 2 sin^{4} (x) + 2 - 4 sin^{2} (x) + 2 sin^{4} (x) : - 4 sin^{2} (x) + 1

- 1 - 2 sin^{4} (x) + 2 - 4 sin^{2} (x) + 2 sin^{4} (x)

条件のようなグループ

= - 2 sin^{4} (x) - 4 sin^{2} (x) + 2 sin^{4} (x) - 1 + 2

類似した元を足す:

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

= - 4 sin^{2} (x) - 1 + 2

数を足す/引く:

- 1 + 2 = 1

= - 4 sin^{2} (x) + 1

= - 4 sin^{2} (x) + 1

= - 4 sin^{2} (x) + 1

1 - 4 sin^{2} (x) = 0

置換で解く

1 - 4 sin^{2} (x) = 0

仮定:

sin (x) = u

1 - 4 u^{2} = 0

1 - 4 u^{2} = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

1 - 4 u^{2} = 0

1

を右側に移動します

1 - 4 u^{2} = 0

両辺から

1

を引く

1 - 4 u^{2} - 1 = 0 - 1

簡素化

- 4 u^{2} = - 1

- 4 u^{2} = - 1

以下で両辺を割る

- 4

- 4 u^{2} = - 1

以下で両辺を割る

- 4

\frac{- 4 u ^{2}}{- 4} = \frac{- 1}{- 4}

簡素化

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = \frac{1}{4}, u = - \frac{1}{4}

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

\frac{1}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

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

= \frac{1}{2}

規則を適用

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

簡素化

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

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

= \frac{1}{2}

= - \frac{1}{2}

規則を適用

1 = 1

= - \frac{1}{2}

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

代用を戻す

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2}, sin (x) = - \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

以下の一般解

sin (x) = \frac{1}{2}

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 = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

以下の一般解

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

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 = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

すべての解を組み合わせる

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

グラフ

人気の例

csc((2x)/3)-1=0

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

6tan(x)=tan^2(x),x=0

6 tan (x) = tan^{2} (x), x = 0

5cos^2(x)+3cos(x)-1=0

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

3 cot (x) + 5 = 8

solvefor a,sin(arccos(x))=sqrt(1-x^2)

so l v e f or a, sin (arccos (x)) = 1 - x^{2}