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人気のある三角関数 >

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

解

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

解

x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

cos^{2} (x)

を引く

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

三角関数の公式を使用して書き換える

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

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

簡素化

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

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

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

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

括弧を分配する

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

マイナス・プラスの規則を適用する

- (- a) = a, - (a) = - a

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

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

簡素化

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

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

類似した元を足す:

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

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

数を引く:

2 - 1 = 1

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

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

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

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

置換で解く

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

仮定:

sin (x) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 2, b = - 3, c = 1

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

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

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

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

指数の規則を適用する:

n

が偶数であれば

(- a)^{n} = a^{n}

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

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

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

数を引く:

9 - 8 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

\frac{- ( - 3 ) + 1}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

数を足す:

3 + 1 = 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= 1

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

\frac{- ( - 3 ) - 1}{2 \cdot 2}

規則を適用

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

数を引く:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

二次equationの解:

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

代用を戻す

u = sin (x)

sin (x) = 1, sin (x) = \frac{1}{2}

sin (x) = 1, sin (x) = \frac{1}{2}

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

以下の一般解

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

グラフ

人気の例

4sin^2(θ)-4cos(θ)=1

4 sin^{2} (θ) - 4 cos (θ) = 1

(33)/(sin(x))=(29)/(sin(100))

\frac{33}{sin ( x )} = \frac{29}{sin ( 10 0 ^{\circ} )}

2cos^2(x)-1=sin(x)

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

(\frac{1}{2}) = cos (θ)

sin(θ)= 4/5 ,θ,I,cos(2θ)

sin (θ) = \frac{4}{5}, θ, I, cos (2 θ)