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人気のある >

4sin^2(θ/2)+8cos(θ/2)-7=0

解

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0

解

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

+1

度

θ = 12 0^{\circ} + 72 0^{\circ} n, θ = 60 0^{\circ} + 72 0^{\circ} n

解答ステップ

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0

三角関数の公式を使用して書き換える

- 7 + 4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2})

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)

= - 7 + 4 (1 - cos^{2} (\frac{θ}{2})) + 8 cos (\frac{θ}{2})

簡素化

- 7 + 4 (1 - cos^{2} (\frac{θ}{2})) + 8 cos (\frac{θ}{2}) : 8 cos (\frac{θ}{2}) - 4 cos^{2} (\frac{θ}{2}) - 3

- 7 + 4 (1 - cos^{2} (\frac{θ}{2})) + 8 cos (\frac{θ}{2})

拡張

4 (1 - cos^{2} (\frac{θ}{2})) : 4 - 4 cos^{2} (\frac{θ}{2})

4 (1 - cos^{2} (\frac{θ}{2}))

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (\frac{θ}{2})

= 4 \cdot 1 - 4 cos^{2} (\frac{θ}{2})

数を乗じる:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (\frac{θ}{2})

= - 7 + 4 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2})

数を足す/引く:

- 7 + 4 = - 3

= 8 cos (\frac{θ}{2}) - 4 cos^{2} (\frac{θ}{2}) - 3

= 8 cos (\frac{θ}{2}) - 4 cos^{2} (\frac{θ}{2}) - 3

- 3 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) = 0

置換で解く

- 3 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) = 0

仮定:

cos (\frac{θ}{2}) = u

- 3 - 4 u^{2} + 8 u = 0

- 3 - 4 u^{2} + 8 u = 0 : u = \frac{1}{2}, u = \frac{3}{2}

- 3 - 4 u^{2} + 8 u = 0

標準的な形式で書く

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

- 4 u^{2} + 8 u - 3 = 0

解くとthe二次式

- 4 u^{2} + 8 u - 3 = 0

二次Equationの公式:

次の場合:

a = - 4, b = 8, c = - 3

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

8^{2} - 4 (- 4) (- 3) = 4

8^{2} - 4 (- 4) (- 3)

規則を適用

- (- a) = a

= 8^{2} - 4 \cdot 4 \cdot 3

数を乗じる:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

数を引く:

64 - 48 = 16

= 16

数を因数に分解する:

16 = 4^{2}

= 4^{2}

累乗根の規則を適用する:

4^{2} = 4

= 4

u_{1, 2} = \frac{- 8 \pm 4}{2 ( - 4 )}

解を分離する

u_{1} = \frac{- 8 + 4}{2 ( - 4 )}, u_{2} = \frac{- 8 - 4}{2 ( - 4 )}

u = \frac{- 8 + 4}{2 ( - 4 )} : \frac{1}{2}

\frac{- 8 + 4}{2 ( - 4 )}

括弧を削除する:

(- a) = - a

= \frac{- 8 + 4}{- 2 \cdot 4}

数を足す/引く:

- 8 + 4 = - 4

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

数を乗じる:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

分数の規則を適用する:

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

= \frac{4}{8}

共通因数を約分する:

4

= \frac{1}{2}

u = \frac{- 8 - 4}{2 ( - 4 )} : \frac{3}{2}

\frac{- 8 - 4}{2 ( - 4 )}

括弧を削除する:

(- a) = - a

= \frac{- 8 - 4}{- 2 \cdot 4}

数を引く:

- 8 - 4 = - 12

= \frac{- 12}{- 2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{- 12}{- 8}

分数の規則を適用する:

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

= \frac{12}{8}

共通因数を約分する:

4

= \frac{3}{2}

二次equationの解:

u = \frac{1}{2}, u = \frac{3}{2}

代用を戻す

u = cos (\frac{θ}{2})

cos (\frac{θ}{2}) = \frac{1}{2}, cos (\frac{θ}{2}) = \frac{3}{2}

cos (\frac{θ}{2}) = \frac{1}{2}, cos (\frac{θ}{2}) = \frac{3}{2}

cos (\frac{θ}{2}) = \frac{1}{2} : θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

cos (\frac{θ}{2}) = \frac{1}{2}

以下の一般解

cos (\frac{θ}{2}) = \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}

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

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

解く

\frac{θ}{2} = \frac{π}{3} + 2 πn : θ = \frac{2 π}{3} + 4 πn

\frac{θ}{2} = \frac{π}{3} + 2 πn

以下で両辺を乗じる:

2

\frac{θ}{2} = \frac{π}{3} + 2 πn

以下で両辺を乗じる:

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

簡素化

\frac{2 θ}{2} = 2 \cdot \frac{π}{3} + 2 \cdot 2 πn

簡素化

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数を割る:

\frac{2}{2} = 1

= θ

簡素化

2 \cdot \frac{π}{3} + 2 \cdot 2 πn : \frac{2 π}{3} + 4 πn

2 \cdot \frac{π}{3} + 2 \cdot 2 πn

乗じる

2 \cdot \frac{π}{3} : \frac{2 π}{3}

2 \cdot \frac{π}{3}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{3}

= \frac{2 π}{3} + 2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn

解く

\frac{θ}{2} = \frac{5 π}{3} + 2 πn : θ = \frac{10 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

以下で両辺を乗じる:

2

\frac{θ}{2} = \frac{5 π}{3} + 2 πn

以下で両辺を乗じる:

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

簡素化

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

簡素化

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数を割る:

\frac{2}{2} = 1

= θ

簡素化

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn : \frac{10 π}{3} + 4 πn

2 \cdot \frac{5 π}{3} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{3} = \frac{10 π}{3}

2 \cdot \frac{5 π}{3}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{5 π 2}{3}

数を乗じる:

5 \cdot 2 = 10

= \frac{10 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

数を乗じる:

2 \cdot 2 = 4

= 4 πn

= \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{10 π}{3} + 4 πn

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

cos (\frac{θ}{2}) = \frac{3}{2} :

解なし

cos (\frac{θ}{2}) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

解なし

すべての解を組み合わせる

θ = \frac{2 π}{3} + 4 πn, θ = \frac{10 π}{3} + 4 πn

人気の例

11sin(x)+5=sin(x)

11 sin (x) + 5 = sin (x)

cos (x) = 2 sin (x)

- 16 sin (4 x) = 0

3 tan (θ) + 1 = 0

2tan^2(x)=3sec(x)

2 tan^{2} (x) = 3 sec (x)