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人気のある三角関数 >

2cos(2x)+sin(x)= 1/2

解

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

解

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn

+1

度

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n, x = 48.59037 \dots^{\circ} + 36 0^{\circ} n, x = 131.40962 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

\frac{1}{2}

を引く

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

簡素化

2 cos (2 x) + sin (x) - \frac{1}{2} : \frac{4 cos ( 2 x ) + 2 sin ( x ) - 1}{2}

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

元を分数に変換する:

2 cos (2 x) = \frac{2 cos ( 2 x ) 2}{2}, sin (x) = \frac{sin ( x ) 2}{2}

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

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

数を乗じる:

2 \cdot 2 = 4

= \frac{4 cos ( 2 x ) + 2 sin ( x ) - 1}{2}

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

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

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

三角関数の公式を使用して書き換える

- 1 + 2 sin (x) + 4 cos (2 x)

2倍角の公式を使用:

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

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

簡素化

- 1 + 2 sin (x) + 4 (1 - 2 sin^{2} (x)) : 2 sin (x) - 8 sin^{2} (x) + 3

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

拡張

4 (1 - 2 sin^{2} (x)) : 4 - 8 sin^{2} (x)

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

簡素化

4 \cdot 1 - 4 \cdot 2 sin^{2} (x) : 4 - 8 sin^{2} (x)

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

数を乗じる:

4 \cdot 1 = 4

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

数を乗じる:

4 \cdot 2 = 8

= 4 - 8 sin^{2} (x)

= 4 - 8 sin^{2} (x)

= - 1 + 2 sin (x) + 4 - 8 sin^{2} (x)

簡素化

- 1 + 2 sin (x) + 4 - 8 sin^{2} (x) : 2 sin (x) - 8 sin^{2} (x) + 3

- 1 + 2 sin (x) + 4 - 8 sin^{2} (x)

条件のようなグループ

= 2 sin (x) - 8 sin^{2} (x) - 1 + 4

数を足す/引く:

- 1 + 4 = 3

= 2 sin (x) - 8 sin^{2} (x) + 3

= 2 sin (x) - 8 sin^{2} (x) + 3

= 2 sin (x) - 8 sin^{2} (x) + 3

3 + 2 sin (x) - 8 sin^{2} (x) = 0

置換で解く

3 + 2 sin (x) - 8 sin^{2} (x) = 0

仮定:

sin (x) = u

3 + 2 u - 8 u^{2} = 0

3 + 2 u - 8 u^{2} = 0 : u = - \frac{1}{2}, u = \frac{3}{4}

3 + 2 u - 8 u^{2} = 0

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 8, b = 2, c = 3

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 8 ) \cdot 3}{2 ( - 8 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 8 ) \cdot 3}{2 ( - 8 )}

2^{2} - 4 (- 8) \cdot 3 = 10

2^{2} - 4 (- 8) \cdot 3

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 8 \cdot 3

数を乗じる:

4 \cdot 8 \cdot 3 = 96

= 2^{2} + 96

2^{2} = 4

= 4 + 96

数を足す:

4 + 96 = 100

= 100

数を因数に分解する:

100 = 1 0^{2}

= 1 0^{2}

累乗根の規則を適用する:

n a^{n} = a

1 0^{2} = 10

= 10

u_{1, 2} = \frac{- 2 \pm 10}{2 ( - 8 )}

解を分離する

u_{1} = \frac{- 2 + 10}{2 ( - 8 )}, u_{2} = \frac{- 2 - 10}{2 ( - 8 )}

u = \frac{- 2 + 10}{2 ( - 8 )} : - \frac{1}{2}

\frac{- 2 + 10}{2 ( - 8 )}

括弧を削除する:

(- a) = - a

= \frac{- 2 + 10}{- 2 \cdot 8}

数を足す/引く:

- 2 + 10 = 8

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

数を乗じる:

2 \cdot 8 = 16

= \frac{8}{- 16}

分数の規則を適用する:

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

= - \frac{8}{16}

共通因数を約分する:

8

= - \frac{1}{2}

u = \frac{- 2 - 10}{2 ( - 8 )} : \frac{3}{4}

\frac{- 2 - 10}{2 ( - 8 )}

括弧を削除する:

(- a) = - a

= \frac{- 2 - 10}{- 2 \cdot 8}

数を引く:

- 2 - 10 = - 12

= \frac{- 12}{- 2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{- 12}{- 16}

分数の規則を適用する:

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

= \frac{12}{16}

共通因数を約分する:

4

= \frac{3}{4}

二次equationの解:

u = - \frac{1}{2}, u = \frac{3}{4}

代用を戻す

u = sin (x)

sin (x) = - \frac{1}{2}, sin (x) = \frac{3}{4}

sin (x) = - \frac{1}{2}, sin (x) = \frac{3}{4}

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

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

sin (x) = \frac{3}{4}

三角関数の逆数プロパティを適用する

sin (x) = \frac{3}{4}

以下の一般解

sin (x) = \frac{3}{4}

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

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

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

すべての解を組み合わせる

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = arcsin (\frac{3}{4}) + 2 πn, x = π - arcsin (\frac{3}{4}) + 2 πn

10進法形式で解を証明する

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn, x = 0.84806 \dots + 2 πn, x = π - 0.84806 \dots + 2 πn

グラフ

人気の例

sqrt(2)sin(2x)=sin(4x),0<= x<= pi

2 sin (2 x) = sin (4 x), 0 \leq x \leq π

9sin(x)-2=4sin^2(x)

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

-2cot(x)=csc^2(x)

- 2 cot (x) = csc^{2} (x)

sin (θ) = - 0.671

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

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