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人気のある三角関数 >

3cos^2(x)+2cos(x)-1=0

解

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

解

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = π + 2 πn

+1

度

x = 70.52877 \dots^{\circ} + 36 0^{\circ} n, x = 289.47122 \dots^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

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

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

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

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

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

規則を適用

- (- a) = a

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

数を乗じる:

4 \cdot 3 \cdot 1 = 12

= 2^{2} + 12

2^{2} = 4

= 4 + 12

数を足す:

4 + 12 = 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:

n a^{n} = a

4^{2} = 4

= 4

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

解を分離する

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

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

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

数を足す/引く:

- 2 + 4 = 2

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

数を乗じる:

2 \cdot 3 = 6

= \frac{2}{6}

共通因数を約分する:

2

= \frac{1}{3}

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

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

数を引く:

- 2 - 4 = - 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 = cos (x)

cos (x) = \frac{1}{3}, cos (x) = - 1

cos (x) = \frac{1}{3}, cos (x) = - 1

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

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

三角関数の逆数プロパティを適用する

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

以下の一般解

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

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

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

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

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

以下の一般解

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 1.23095 \dots + 2 πn, x = 2 π - 1.23095 \dots + 2 πn, x = π + 2 πn

グラフ

人気の例

arctan (θ) = - 2

sin(3x)-sin(x)=sin(3x-x)

sin (3 x) - sin (x) = sin (3 x - x)

sin(a)=-(sqrt(3))/2

sin (a) = - \frac{3}{2}

solvefor t,v=311cos(30t+30)v

so l v e f or t, v = 311 cos (30 t + 3 0^{\circ}) v

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

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