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人気のある三角関数 >

sin(2x)+cos(3x)=0

解

sin (2 x) + cos (3 x) = 0

解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = - 1 8^{\circ} + 36 0^{\circ} n, x = 19 8^{\circ} + 36 0^{\circ} n, x = 5 4^{\circ} + 36 0^{\circ} n, x = 12 6^{\circ} + 36 0^{\circ} n

解答ステップ

sin (2 x) + cos (3 x) = 0

三角関数の公式を使用して書き換える

cos (3 x) + sin (2 x)

2倍角の公式を使用:

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

= cos (3 x) + 2 sin (x) cos (x)

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

cos (3 x)

三角関数の公式を使用して書き換える

cos (3 x)

書き換え

= cos (2 x + x)

角の和の公式を使用する:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

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

2倍角の公式を使用:

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

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

簡素化

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

cos (2 x) cos (x) - 2 sin (x) cos (x) sin (x)

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

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

指数の規則を適用する:

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

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

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

数を足す:

1 + 1 = 2

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

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

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

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

2倍角の公式を使用:

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

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

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

拡張

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

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

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

簡素化

2 cos^{2} (x) cos (x) - 1 \cdot cos (x) : 2 cos^{3} (x) - cos (x)

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

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

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

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 cos^{3} (x)

1 \cdot cos (x) = cos (x)

1 cos (x)

乗算:

1 \cdot cos (x) = cos (x)

= cos (x)

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

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

マイナス・プラスの規則を適用する

- (- a) = a

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

簡素化

- 2 \cdot 1 \cdot cos (x) + 2 cos^{2} (x) cos (x) : - 2 cos (x) + 2 cos^{3} (x)

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

2 \cdot 1 \cdot cos (x) = 2 cos (x)

2 \cdot 1 cos (x)

数を乗じる:

2 \cdot 1 = 2

= 2 cos (x)

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

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

指数の規則を適用する:

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

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

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

数を足す:

2 + 1 = 3

= 2 cos^{3} (x)

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

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

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

類似した元を足す:

- cos (x) - 2 cos (x) = - 3 cos (x)

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

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

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

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

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

因数

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

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

指数の規則を適用する:

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

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

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

共通項をくくり出す

cos (x)

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

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

各部分を別個に解く

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

以下の一般解

cos (x) = 0

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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

- 3 + 4 cos^{2} (x) + 2 sin (x) = 0 : x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

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

三角関数の公式を使用して書き換える

- 3 + 2 sin (x) + 4 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)

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

4 \cdot 1 = 4

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

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

簡素化

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

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

条件のようなグループ

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

数を足す/引く:

- 3 + 4 = 1

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

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

1 + 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 = 0

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

2^{2} - 4 (- 4) \cdot 1 = 25

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

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 1

数を乗じる:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

数を足す:

4 + 16 = 20

= 20

以下の素因数分解:

20 : 2^{2} \cdot 5

20

20220 = 10 \cdot 2

で割る

= 2 \cdot 10

10210 = 5 \cdot 2

で割る

= 2 \cdot 2 \cdot 5

2, 5

はすべて素数である。ゆえにさらに因数分解することはできない

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

累乗根の規則を適用する:

n ab = n a n b

= 5 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 25

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

解を分離する

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

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

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

括弧を削除する:

(- a) = - a

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

数を乗じる:

2 \cdot 4 = 8

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

分数の規則を適用する:

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

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

キャンセル

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

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

因数

- 2 + 25 : 2 (- 1 + 5)

- 2 + 25

書き換え

= - 2 \cdot 1 + 25

共通項をくくり出す

2

= 2 (- 1 + 5)

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

共通因数を約分する:

2

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

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

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

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

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

括弧を削除する:

(- a) = - a

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

数を乗じる:

2 \cdot 4 = 8

= \frac{- 2 - 2 5}{- 8}

分数の規則を適用する:

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

- 2 - 25 = - (2 + 25)

= \frac{2 + 2 5}{8}

因数

2 + 25 : 2 (1 + 5)

2 + 25

書き換え

= 2 \cdot 1 + 25

共通項をくくり出す

2

= 2 (1 + 5)

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

共通因数を約分する:

2

= \frac{1 + 5}{4}

二次equationの解:

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

代用を戻す

u = sin (x)

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

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

sin (x) = - \frac{- 1 + 5}{4} : x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

sin (x) = \frac{1 + 5}{4} : x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

すべての解を組み合わせる

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

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arcsin (- \frac{- 1 + 5}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

10進法形式で解を証明する

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn

グラフ

人気の例

2cos^2(x)+2cos(x)-1=0,0<= x<= 2pi

2 cos^{2} (x) + 2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

sec (x) = \frac{7}{5}

cos(5θ)=sin(3θ)

cos (5 θ) = sin (3 θ)

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

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

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