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人気のある三角関数 >

sin^2(x)+cos(x)= 5/4

解

sin^{2} (x) + cos (x) = \frac{5}{4}

解

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

+1

度

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

解答ステップ

sin^{2} (x) + cos (x) = \frac{5}{4}

両辺から

\frac{5}{4}

を引く

sin^{2} (x) + cos (x) - \frac{5}{4} = 0

簡素化

sin^{2} (x) + cos (x) - \frac{5}{4} : \frac{4 sin ^{2} ( x ) + 4 cos ( x ) - 5}{4}

sin^{2} (x) + cos (x) - \frac{5}{4}

元を分数に変換する:

sin^{2} (x) = \frac{sin ^{2} ( x ) 4}{4}, cos (x) = \frac{cos ( x ) 4}{4}

= \frac{sin ^{2} ( x ) \cdot 4}{4} + \frac{cos ( x ) \cdot 4}{4} - \frac{5}{4}

分母が等しいので, 分数を組み合わせる:

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

= \frac{sin ^{2} ( x ) \cdot 4 + cos ( x ) \cdot 4 - 5}{4}

\frac{4 sin ^{2} ( x ) + 4 cos ( x ) - 5}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin^{2} (x) + 4 cos (x) - 5 = 0

三角関数の公式を使用して書き換える

- 5 + 4 cos (x) + 4 sin^{2} (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)

= - 5 + 4 cos (x) + 4 (1 - cos^{2} (x))

簡素化

- 5 + 4 cos (x) + 4 (1 - cos^{2} (x)) : 4 cos (x) - 4 cos^{2} (x) - 1

- 5 + 4 cos (x) + 4 (1 - cos^{2} (x))

拡張

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

数を乗じる:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (x)

= - 5 + 4 cos (x) + 4 - 4 cos^{2} (x)

簡素化

- 5 + 4 cos (x) + 4 - 4 cos^{2} (x) : 4 cos (x) - 4 cos^{2} (x) - 1

- 5 + 4 cos (x) + 4 - 4 cos^{2} (x)

条件のようなグループ

= 4 cos (x) - 4 cos^{2} (x) - 5 + 4

数を足す/引く:

- 5 + 4 = - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

= 4 cos (x) - 4 cos^{2} (x) - 1

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

4^{2} - 4 (- 4) (- 1) = 0

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

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

数を引く:

16 - 16 = 0

= 0

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

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

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

括弧を削除する:

(- a) = - a

= \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{1}{2}

二次equationの解:

u = \frac{1}{2}

代用を戻す

u = cos (x)

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

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

グラフ

人気の例

cos(2θ)+2=5sin(θ)

tan(x)-sin(2x)=0

sin(x)-cos(x)+1=0