ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある三角関数 >

3cos^2(x)+4sin(x)-4=0

解

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

解

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn, x = \frac{π}{2} + 2 πn

+1

度

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

三角関数の公式を使用して書き換える

- 4 + 3 cos^{2} (x) + 4 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)

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

簡素化

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

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

拡張

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

3 (1 - sin^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (x)

= 3 \cdot 1 - 3 sin^{2} (x)

数を乗じる:

3 \cdot 1 = 3

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

= - 4 + 3 - 3 sin^{2} (x) + 4 sin (x)

数を足す/引く:

- 4 + 3 = - 1

= 4 sin (x) - 3 sin^{2} (x) - 1

= 4 sin (x) - 3 sin^{2} (x) - 1

- 1 - 3 sin^{2} (x) + 4 sin (x) = 0

置換で解く

- 1 - 3 sin^{2} (x) + 4 sin (x) = 0

仮定:

sin (x) = u

- 1 - 3 u^{2} + 4 u = 0

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

- 1 - 3 u^{2} + 4 u = 0

標準的な形式で書く

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

- 3 u^{2} + 4 u - 1 = 0

解くとthe二次式

- 3 u^{2} + 4 u - 1 = 0

二次Equationの公式:

次の場合:

a = - 3, b = 4, c = - 1

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

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

4^{2} - 4 (- 3) (- 1) = 2

4^{2} - 4 (- 3) (- 1)

規則を適用

- (- a) = a

= 4^{2} - 4 \cdot 3 \cdot 1

数を乗じる:

4 \cdot 3 \cdot 1 = 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

数を引く:

16 - 12 = 4

= 4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

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

解を分離する

u_{1} = \frac{- 4 + 2}{2 ( - 3 )}, u_{2} = \frac{- 4 - 2}{2 ( - 3 )}

u = \frac{- 4 + 2}{2 ( - 3 )} : \frac{1}{3}

\frac{- 4 + 2}{2 ( - 3 )}

括弧を削除する:

(- a) = - a

= \frac{- 4 + 2}{- 2 \cdot 3}

数を足す/引く:

- 4 + 2 = - 2

= \frac{- 2}{- 2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{- 2}{- 6}

分数の規則を適用する:

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

= \frac{2}{6}

共通因数を約分する:

2

= \frac{1}{3}

u = \frac{- 4 - 2}{2 ( - 3 )} : 1

\frac{- 4 - 2}{2 ( - 3 )}

括弧を削除する:

(- a) = - a

= \frac{- 4 - 2}{- 2 \cdot 3}

数を引く:

- 4 - 2 = - 6

= \frac{- 6}{- 2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{- 6}{- 6}

分数の規則を適用する:

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

= \frac{6}{6}

規則を適用

\frac{a}{a} = 1

= 1

二次equationの解:

u = \frac{1}{3}, u = 1

代用を戻す

u = sin (x)

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

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

sin (x) = \frac{1}{3} : x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

sin (x) = \frac{1}{3}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1}{3}

以下の一般解

sin (x) = \frac{1}{3}

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

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

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

すべての解を組み合わせる

x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn, x = \frac{π}{2} + 2 πn

10進法形式で解を証明する

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn, x = \frac{π}{2} + 2 πn

グラフ

人気の例

sin (x) = \frac{7}{15}

-4sin(x)=-cos^2(x)+4,0<= x<= 2pi

- 4 sin (x) = - cos^{2} (x) + 4, 0 \leq x \leq 2 π

sin(2x)cot(x)=0

sin (2 x) cot (x) = 0

3sec(θ)-2sqrt(3)=0

3 sec (θ) - 23 = 0

sqrt(3)sin(2x)+cos(2x)=2cos(2x)

3 sin (2 x) + cos (2 x) = 2 cos (2 x)