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人気のある三角関数 >

2sin^2(x)+9cos(x)-6=0

解

2 sin^{2} (x) + 9 cos (x) - 6 = 0

解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

+1

度

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin^{2} (x) + 9 cos (x) - 6 = 0

三角関数の公式を使用して書き換える

- 6 + 2 sin^{2} (x) + 9 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)

= - 6 + 2 (1 - cos^{2} (x)) + 9 cos (x)

簡素化

- 6 + 2 (1 - cos^{2} (x)) + 9 cos (x) : 9 cos (x) - 2 cos^{2} (x) - 4

- 6 + 2 (1 - cos^{2} (x)) + 9 cos (x)

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

= - 6 + 2 - 2 cos^{2} (x) + 9 cos (x)

数を足す/引く:

- 6 + 2 = - 4

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

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

- 4 - 2 cos^{2} (x) + 9 cos (x) = 0

置換で解く

- 4 - 2 cos^{2} (x) + 9 cos (x) = 0

仮定:

cos (x) = u

- 4 - 2 u^{2} + 9 u = 0

- 4 - 2 u^{2} + 9 u = 0 : u = \frac{1}{2}, u = 4

- 4 - 2 u^{2} + 9 u = 0

標準的な形式で書く

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

- 2 u^{2} + 9 u - 4 = 0

解くとthe二次式

- 2 u^{2} + 9 u - 4 = 0

二次Equationの公式:

次の場合:

a = - 2, b = 9, c = - 4

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 2 ) ( - 4 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 ( - 2 ) ( - 4 )}{2 ( - 2 )}

9^{2} - 4 (- 2) (- 4) = 7

9^{2} - 4 (- 2) (- 4)

規則を適用

- (- a) = a

= 9^{2} - 4 \cdot 2 \cdot 4

数を乗じる:

4 \cdot 2 \cdot 4 = 32

= 9^{2} - 32

9^{2} = 81

= 81 - 32

数を引く:

81 - 32 = 49

= 49

数を因数に分解する:

49 = 7^{2}

= 7^{2}

累乗根の規則を適用する:

n a^{n} = a

7^{2} = 7

= 7

u_{1, 2} = \frac{- 9 \pm 7}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 9 + 7}{2 ( - 2 )}, u_{2} = \frac{- 9 - 7}{2 ( - 2 )}

u = \frac{- 9 + 7}{2 ( - 2 )} : \frac{1}{2}

\frac{- 9 + 7}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 9 + 7}{- 2 \cdot 2}

数を足す/引く:

- 9 + 7 = - 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

分数の規則を適用する:

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

u = \frac{- 9 - 7}{2 ( - 2 )} : 4

\frac{- 9 - 7}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 9 - 7}{- 2 \cdot 2}

数を引く:

- 9 - 7 = - 16

= \frac{- 16}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 16}{- 4}

分数の規則を適用する:

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

= \frac{16}{4}

数を割る:

\frac{16}{4} = 4

= 4

二次equationの解:

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

代用を戻す

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = 4

cos (x) = \frac{1}{2}, cos (x) = 4

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

以下の一般解

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = 4 :

解なし

cos (x) = 4

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

グラフ

人気の例

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

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

solvefor y,x=sin(2y)

so l v e f or y, x = sin (2 y)

3cos(θ)=3sin(θ)

3 cos (θ) = 3 sin (θ)

sin(x+pi/3)+sin(x-pi/3)=(sqrt(3))/2

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

sin (A) = \frac{12}{13}