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人気のある三角関数 >

4sin^2(x)+2cos(x)+a=3

解

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

解

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

解答ステップ

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

両辺から

3

を引く

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

三角関数の公式を使用して書き換える

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

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)

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

簡素化

- 3 + a + 2 cos (x) + 4 (1 - cos^{2} (x)) : 2 cos (x) - 4 cos^{2} (x) + a + 1

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

拡張

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

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

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

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

数を乗じる:

4 \cdot 1 = 4

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

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

簡素化

- 3 + a + 2 cos (x) + 4 - 4 cos^{2} (x) : 2 cos (x) - 4 cos^{2} (x) + a + 1

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

条件のようなグループ

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

数を足す/引く:

- 3 + 4 = 1

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

1 + a + 2 u - 4 u^{2} = 0 : u = - \frac{- 1 + 4 a + 5}{4}, u = \frac{4 a + 5 + 1}{4}

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 4, b = 2, c = 1 + a

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 4 ) ( 1 + a )}{2 ( - 4 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 4 ) ( 1 + a )}{2 ( - 4 )}

簡素化

2^{2} - 4 (- 4) (1 + a) : 2 4 a + 5

2^{2} - 4 (- 4) (1 + a)

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 4 = 16

= 2^{2} + 16 (a + 1)

因数

2^{2} + 16 (1 + a) : 4 (4 a + 5)

2^{2} + 16 (1 + a)

書き換え

= 4 \cdot 1 + 4 \cdot 4 (1 + a)

共通項をくくり出す

4

= 4 (1 + 4 (1 + a))

拡張

4 (a + 1) + 1 : 4 a + 5

1 + 4 (1 + a)

拡張

4 (1 + a) : 4 + 4 a

4 (1 + a)

分配法則を適用する:

a (b + c) = ab + a c

a = 4, b = 1, c = a

= 4 \cdot 1 + 4 a

数を乗じる:

4 \cdot 1 = 4

= 4 + 4 a

= 1 + 4 + 4 a

数を足す:

1 + 4 = 5

= 4 a + 5

= 4 (4 a + 5)

= 4 (4 a + 5)

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= 4 4 a + 5

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

= 2 4 a + 5

u_{1, 2} = \frac{- 2 \pm 2 4 a + 5}{2 ( - 4 )}

解を分離する

u_{1} = \frac{- 2 + 2 4 a + 5}{2 ( - 4 )}, u_{2} = \frac{- 2 - 2 4 a + 5}{2 ( - 4 )}

u = \frac{- 2 + 2 4 a + 5}{2 ( - 4 )} : - \frac{- 1 + 4 a + 5}{4}

\frac{- 2 + 2 4 a + 5}{2 ( - 4 )}

括弧を削除する:

(- a) = - a

= \frac{- 2 + 2 4 a + 5}{- 2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 2 + 2 4 a + 5}{- 8}

分数の規則を適用する:

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

= - \frac{- 2 + 2 4 a + 5}{8}

キャンセル

\frac{- 2 + 2 4 a + 5}{8} : \frac{4 a + 5 - 1}{4}

\frac{- 2 + 2 4 a + 5}{8}

因数

- 2 + 2 4 a + 5 : 2 (- 1 + 5 + 4 a)

- 2 + 2 4 a + 5

書き換え

= - 2 \cdot 1 + 2 5 + 4 a

共通項をくくり出す

2

= 2 (- 1 + 5 + 4 a)

= \frac{2 ( - 1 + 5 + 4 a )}{8}

共通因数を約分する:

2

= \frac{- 1 + 4 a + 5}{4}

= - \frac{4 a + 5 - 1}{4}

= - \frac{- 1 + 4 a + 5}{4}

u = \frac{- 2 - 2 4 a + 5}{2 ( - 4 )} : \frac{4 a + 5 + 1}{4}

\frac{- 2 - 2 4 a + 5}{2 ( - 4 )}

括弧を削除する:

(- a) = - a

= \frac{- 2 - 2 4 a + 5}{- 2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 2 - 2 4 a + 5}{- 8}

分数の規則を適用する:

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

= - \frac{- 2 - 2 4 a + 5}{8}

キャンセル

\frac{- 2 - 2 4 a + 5}{8} : - \frac{4 a + 5 + 1}{4}

\frac{- 2 - 2 4 a + 5}{8}

因数

- 2 - 2 4 a + 5 : - 2 (1 + 5 + 4 a)

- 2 - 2 4 a + 5

書き換え

= - 2 \cdot 1 - 2 5 + 4 a

共通項をくくり出す

2

= - 2 (1 + 5 + 4 a)

= - \frac{2 ( 1 + 5 + 4 a )}{8}

共通因数を約分する:

2

= - \frac{4 a + 5 + 1}{4}

= - (- \frac{4 a + 5 + 1}{4})

規則を適用

- (- a) = a

= \frac{4 a + 5 + 1}{4}

二次equationの解:

u = - \frac{- 1 + 4 a + 5}{4}, u = \frac{4 a + 5 + 1}{4}

代用を戻す

u = cos (x)

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4}, cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = - \frac{- 1 + 4 a + 5}{4} : x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

cos (x) = - \frac{- 1 + 4 a + 5}{4}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{- 1 + 4 a + 5}{4}

以下の一般解

cos (x) = - \frac{- 1 + 4 a + 5}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn

cos (x) = \frac{4 a + 5 + 1}{4} : x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

cos (x) = \frac{4 a + 5 + 1}{4}

三角関数の逆数プロパティを適用する

cos (x) = \frac{4 a + 5 + 1}{4}

以下の一般解

cos (x) = \frac{4 a + 5 + 1}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

すべての解を組み合わせる

x = arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = - arccos (- \frac{- 1 + 4 a + 5}{4}) + 2 πn, x = arccos (\frac{4 a + 5 + 1}{4}) + 2 πn, x = - arccos (\frac{4 a + 5 + 1}{4}) + 2 πn

グラフ

人気の例

cos^5(x)+cos(x)+4cos^2(x)=2

cos^{5} (x) + cos (x) + 4 cos^{2} (x) = 2

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

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

(sin^3(x))/(2+2(sin(x))^2)=1

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

solvefor x,cot^2(x)= 1/3

so l v e f or x, cot^{2} (x) = \frac{1}{3}

2+cos^2(x)=-5sin(x)

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