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人気のある三角関数 >

10sin(x)+10cos^2(x)=10,0<= x<2pi

解

10 sin (x) + 10 cos^{2} (x) = 10, 0 \leq x < 2 π

解

x = 0, x = π, x = \frac{π}{2}

+1

度

x = 0^{\circ}, x = 18 0^{\circ}, x = 9 0^{\circ}

解答ステップ

10 sin (x) + 10 cos^{2} (x) = 10, 0 \leq x < 2 π

両辺から

10

を引く

10 sin (x) + 10 cos^{2} (x) - 10 = 0

三角関数の公式を使用して書き換える

- 10 + 10 cos^{2} (x) + 10 sin (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)

= - 10 + 10 (1 - sin^{2} (x)) + 10 sin (x)

簡素化

- 10 + 10 (1 - sin^{2} (x)) + 10 sin (x) : 10 sin (x) - 10 sin^{2} (x)

- 10 + 10 (1 - sin^{2} (x)) + 10 sin (x)

拡張

10 (1 - sin^{2} (x)) : 10 - 10 sin^{2} (x)

10 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 10, b = 1, c = sin^{2} (x)

= 10 \cdot 1 - 10 sin^{2} (x)

数を乗じる:

10 \cdot 1 = 10

= 10 - 10 sin^{2} (x)

= - 10 + 10 - 10 sin^{2} (x) + 10 sin (x)

- 10 + 10 = 0

= 10 sin (x) - 10 sin^{2} (x)

= 10 sin (x) - 10 sin^{2} (x)

10 sin (x) - 10 sin^{2} (x) = 0

置換で解く

10 sin (x) - 10 sin^{2} (x) = 0

仮定:

sin (x) = u

10 u - 10 u^{2} = 0

10 u - 10 u^{2} = 0 : u = 0, u = 1

10 u - 10 u^{2} = 0

標準的な形式で書く

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

- 10 u^{2} + 10 u = 0

解くとthe二次式

- 10 u^{2} + 10 u = 0

二次Equationの公式:

次の場合:

a = - 10, b = 10, c = 0

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

u_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 10 ) \cdot 0}{2 ( - 10 )}

1 0^{2} - 4 (- 10) \cdot 0 = 10

1 0^{2} - 4 (- 10) \cdot 0

規則を適用

- (- a) = a

= 1 0^{2} + 4 \cdot 10 \cdot 0

規則を適用

0 \cdot a = 0

= 1 0^{2} + 0

1 0^{2} + 0 = 1 0^{2}

= 1 0^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 10

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

解を分離する

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

u = \frac{- 10 + 10}{2 ( - 10 )} : 0

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

括弧を削除する:

(- a) = - a

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

数を足す/引く:

- 10 + 10 = 0

= \frac{0}{- 2 \cdot 10}

数を乗じる:

2 \cdot 10 = 20

= \frac{0}{- 20}

分数の規則を適用する:

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

= - \frac{0}{20}

規則を適用

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

= - 0

= 0

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

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

括弧を削除する:

(- a) = - a

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

数を引く:

- 10 - 10 = - 20

= \frac{- 20}{- 2 \cdot 10}

数を乗じる:

2 \cdot 10 = 20

= \frac{- 20}{- 20}

分数の規則を適用する:

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

= \frac{20}{20}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

u = 0, u = 1

代用を戻す

u = sin (x)

sin (x) = 0, sin (x) = 1

sin (x) = 0, sin (x) = 1

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

以下の一般解

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

範囲の解答

0 \leq x < 2 π

x = 0, x = π

sin (x) = 1, 0 \leq x < 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x < 2 π

以下の一般解

sin (x) = 1

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}{2} + 2 πn

x = \frac{π}{2} + 2 πn

範囲の解答

0 \leq x < 2 π

x = \frac{π}{2}

すべての解を組み合わせる

x = 0, x = π, x = \frac{π}{2}

グラフ

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