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人気のある三角関数 >

2sin^2(x)+5cos(x)-3=0

解

2 sin^{2} (x) + 5 cos (x) - 3 = 0

解

x = 1.34977 \dots + 2 πn, x = 2 π - 1.34977 \dots + 2 πn

+1

度

x = 77.33656 \dots^{\circ} + 36 0^{\circ} n, x = 282.66343 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

2 sin^{2} (x) + 5 cos (x) - 3 = 0

三角関数の公式を使用して書き換える

- 3 + 2 sin^{2} (x) + 5 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)

= - 3 + 2 (1 - cos^{2} (x)) + 5 cos (x)

簡素化

- 3 + 2 (1 - cos^{2} (x)) + 5 cos (x) : 5 cos (x) - 2 cos^{2} (x) - 1

- 3 + 2 (1 - cos^{2} (x)) + 5 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)

= - 3 + 2 - 2 cos^{2} (x) + 5 cos (x)

数を足す/引く:

- 3 + 2 = - 1

= 5 cos (x) - 2 cos^{2} (x) - 1

= 5 cos (x) - 2 cos^{2} (x) - 1

- 1 - 2 cos^{2} (x) + 5 cos (x) = 0

置換で解く

- 1 - 2 cos^{2} (x) + 5 cos (x) = 0

仮定:

cos (x) = u

- 1 - 2 u^{2} + 5 u = 0

- 1 - 2 u^{2} + 5 u = 0 : u = \frac{5 - 17}{4}, u = \frac{5 + 17}{4}

- 1 - 2 u^{2} + 5 u = 0

標準的な形式で書く

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

- 2 u^{2} + 5 u - 1 = 0

解くとthe二次式

- 2 u^{2} + 5 u - 1 = 0

二次Equationの公式:

次の場合:

a = - 2, b = 5, c = - 1

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

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

5^{2} - 4 (- 2) (- 1) = 17

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

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 5^{2} - 8

5^{2} = 25

= 25 - 8

数を引く:

25 - 8 = 17

= 17

u_{1, 2} = \frac{- 5 \pm 17}{2 ( - 2 )}

解を分離する

u_{1} = \frac{- 5 + 17}{2 ( - 2 )}, u_{2} = \frac{- 5 - 17}{2 ( - 2 )}

u = \frac{- 5 + 17}{2 ( - 2 )} : \frac{5 - 17}{4}

\frac{- 5 + 17}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 + 17}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 5 + 17}{- 4}

分数の規則を適用する:

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

- 5 + 17 = - (5 - 17)

= \frac{5 - 17}{4}

u = \frac{- 5 - 17}{2 ( - 2 )} : \frac{5 + 17}{4}

\frac{- 5 - 17}{2 ( - 2 )}

括弧を削除する:

(- a) = - a

= \frac{- 5 - 17}{- 2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{- 5 - 17}{- 4}

分数の規則を適用する:

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

- 5 - 17 = - (5 + 17)

= \frac{5 + 17}{4}

二次equationの解:

u = \frac{5 - 17}{4}, u = \frac{5 + 17}{4}

代用を戻す

u = cos (x)

cos (x) = \frac{5 - 17}{4}, cos (x) = \frac{5 + 17}{4}

cos (x) = \frac{5 - 17}{4}, cos (x) = \frac{5 + 17}{4}

cos (x) = \frac{5 - 17}{4} : x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

cos (x) = \frac{5 - 17}{4}

三角関数の逆数プロパティを適用する

cos (x) = \frac{5 - 17}{4}

以下の一般解

cos (x) = \frac{5 - 17}{4}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

cos (x) = \frac{5 + 17}{4} :

解なし

cos (x) = \frac{5 + 17}{4}

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

x = arccos (\frac{5 - 17}{4}) + 2 πn, x = 2 π - arccos (\frac{5 - 17}{4}) + 2 πn

10進法形式で解を証明する

x = 1.34977 \dots + 2 πn, x = 2 π - 1.34977 \dots + 2 πn

グラフ

人気の例

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

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

sin(θ)=-27/190

sin (θ) = - \frac{27}{190}

4sin(θ)=1+sin(θ)

4 sin (θ) = 1 + sin (θ)

tan(θ)-sqrt(3)=0,0<= θ<= 2pi

tan (θ) - 3 = 0, 0 \leq θ \leq 2 π

1/32*1200^2*sin(2θ)=3000

\frac{1}{32} \cdot 120 0^{2} \cdot sin (2 θ) = 3000