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人気のある三角関数 >

2cos^2(x)+15sin(x)=9

解

2 cos^{2} (x) + 15 sin (x) = 9

解

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

+1

度

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

解答ステップ

2 cos^{2} (x) + 15 sin (x) = 9

両辺から

9

を引く

2 cos^{2} (x) + 15 sin (x) - 9 = 0

三角関数の公式を使用して書き換える

- 9 + 15 sin (x) + 2 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)

= - 9 + 15 sin (x) + 2 (1 - sin^{2} (x))

簡素化

- 9 + 15 sin (x) + 2 (1 - sin^{2} (x)) : 15 sin (x) - 2 sin^{2} (x) - 7

- 9 + 15 sin (x) + 2 (1 - sin^{2} (x))

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

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

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

数を乗じる:

2 \cdot 1 = 2

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

= - 9 + 15 sin (x) + 2 - 2 sin^{2} (x)

簡素化

- 9 + 15 sin (x) + 2 - 2 sin^{2} (x) : 15 sin (x) - 2 sin^{2} (x) - 7

- 9 + 15 sin (x) + 2 - 2 sin^{2} (x)

条件のようなグループ

= 15 sin (x) - 2 sin^{2} (x) - 9 + 2

数を足す/引く:

- 9 + 2 = - 7

= 15 sin (x) - 2 sin^{2} (x) - 7

= 15 sin (x) - 2 sin^{2} (x) - 7

= 15 sin (x) - 2 sin^{2} (x) - 7

- 7 + 15 sin (x) - 2 sin^{2} (x) = 0

置換で解く

- 7 + 15 sin (x) - 2 sin^{2} (x) = 0

仮定:

sin (x) = u

- 7 + 15 u - 2 u^{2} = 0

- 7 + 15 u - 2 u^{2} = 0 : u = \frac{1}{2}, u = 7

- 7 + 15 u - 2 u^{2} = 0

標準的な形式で書く

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

- 2 u^{2} + 15 u - 7 = 0

解くとthe二次式

- 2 u^{2} + 15 u - 7 = 0

二次Equationの公式:

次の場合:

a = - 2, b = 15, c = - 7

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 ( - 2 ) ( - 7 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 ( - 2 ) ( - 7 )}{2 ( - 2 )}

1 5^{2} - 4 (- 2) (- 7) = 13

1 5^{2} - 4 (- 2) (- 7)

規則を適用

- (- a) = a

= 1 5^{2} - 4 \cdot 2 \cdot 7

数を乗じる:

4 \cdot 2 \cdot 7 = 56

= 1 5^{2} - 56

1 5^{2} = 225

= 225 - 56

数を引く:

225 - 56 = 169

= 169

数を因数に分解する:

169 = 1 3^{2}

= 1 3^{2}

累乗根の規則を適用する:

n a^{n} = a

1 3^{2} = 13

= 13

u_{1, 2} = \frac{- 15 \pm 13}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 15 + 13}{2 ( - 2 )}, u_{2} = \frac{- 15 - 13}{2 ( - 2 )}

u = \frac{- 15 + 13}{2 ( - 2 )} : \frac{1}{2}

\frac{- 15 + 13}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 15 + 13}{- 2 \cdot 2}

数を足す/引く:

- 15 + 13 = - 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{- 15 - 13}{2 ( - 2 )} : 7

\frac{- 15 - 13}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 15 - 13}{- 2 \cdot 2}

数を引く:

- 15 - 13 = - 28

= \frac{- 28}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 28}{- 4}

分数の規則を適用する:

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

= \frac{28}{4}

数を割る:

\frac{28}{4} = 7

= 7

二次equationの解:

u = \frac{1}{2}, u = 7

代用を戻す

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = 7

sin (x) = \frac{1}{2}, sin (x) = 7

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) = 7 :

解なし

sin (x) = 7

- 1 \leq sin (x) \leq 1

解なし

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

グラフ

人気の例

cot (θ) = 1.413

arccos(x)+arccos(2x)=arccos(1/2)

arccos (x) + arccos (2 x) = arccos (\frac{1}{2})

tan(θ)=(2sqrt(3))/3 sin(θ),0<= θ<= 2pi

tan (θ) = \frac{2 3}{3} sin (θ), 0 \leq θ \leq 2 π

solvefor x,cos(x+y)=-cos(x)

so l v e f or x, cos (x + y) = - cos (x)

4+2tan^2(x)=5tan^2(x)+3

4 + 2 tan^{2} (x) = 5 tan^{2} (x) + 3