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人気のある三角関数 >

16sin^2(x)=16-8cos(x)

解

16 sin^{2} (x) = 16 - 8 cos (x)

解

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

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

解答ステップ

16 sin^{2} (x) = 16 - 8 cos (x)

両辺から

16 - 8 cos (x)

を引く

16 sin^{2} (x) - 16 + 8 cos (x) = 0

三角関数の公式を使用して書き換える

- 16 + 16 sin^{2} (x) + 8 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)

= - 16 + 16 (1 - cos^{2} (x)) + 8 cos (x)

簡素化

- 16 + 16 (1 - cos^{2} (x)) + 8 cos (x) : 8 cos (x) - 16 cos^{2} (x)

- 16 + 16 (1 - cos^{2} (x)) + 8 cos (x)

拡張

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

16 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (x)

= 16 \cdot 1 - 16 cos^{2} (x)

数を乗じる:

16 \cdot 1 = 16

= 16 - 16 cos^{2} (x)

= - 16 + 16 - 16 cos^{2} (x) + 8 cos (x)

- 16 + 16 = 0

= 8 cos (x) - 16 cos^{2} (x)

= 8 cos (x) - 16 cos^{2} (x)

- 16 cos^{2} (x) + 8 cos (x) = 0

置換で解く

- 16 cos^{2} (x) + 8 cos (x) = 0

仮定:

cos (x) = u

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

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 16, b = 8, c = 0

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

8^{2} - 4 (- 16) \cdot 0 = 8

8^{2} - 4 (- 16) \cdot 0

規則を適用

- (- a) = a

= 8^{2} + 4 \cdot 16 \cdot 0

規則を適用

0 \cdot a = 0

= 8^{2} + 0

8^{2} + 0 = 8^{2}

= 8^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 8

u_{1, 2} = \frac{- 8 \pm 8}{2 ( - 16 )}

解を分離する

u_{1} = \frac{- 8 + 8}{2 ( - 16 )}, u_{2} = \frac{- 8 - 8}{2 ( - 16 )}

u = \frac{- 8 + 8}{2 ( - 16 )} : 0

\frac{- 8 + 8}{2 ( - 16 )}

括弧を削除する:

(- a) = - a

= \frac{- 8 + 8}{- 2 \cdot 16}

数を足す/引く:

- 8 + 8 = 0

= \frac{0}{- 2 \cdot 16}

数を乗じる:

2 \cdot 16 = 32

= \frac{0}{- 32}

分数の規則を適用する:

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

= - \frac{0}{32}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 8 - 8}{2 ( - 16 )} : \frac{1}{2}

\frac{- 8 - 8}{2 ( - 16 )}

括弧を削除する:

(- a) = - a

= \frac{- 8 - 8}{- 2 \cdot 16}

数を引く:

- 8 - 8 = - 16

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

数を乗じる:

2 \cdot 16 = 32

= \frac{- 16}{- 32}

分数の規則を適用する:

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

= \frac{16}{32}

共通因数を約分する:

16

= \frac{1}{2}

二次equationの解:

u = 0, u = \frac{1}{2}

代用を戻す

u = cos (x)

cos (x) = 0, cos (x) = \frac{1}{2}

cos (x) = 0, cos (x) = \frac{1}{2}

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

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

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

グラフ

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