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人気のある三角関数 >

3cos^2(x)-6cos(x)=-3,0<= x<= 2pi

解

3 cos^{2} (x) - 6 cos (x) = - 3, 0 \leq x \leq 2 π

解

x = 0, x = 2 π

+1

度

x = 0^{\circ}, x = 36 0^{\circ}

解答ステップ

3 cos^{2} (x) - 6 cos (x) = - 3, 0 \leq x \leq 2 π

置換で解く

3 cos^{2} (x) - 6 cos (x) = - 3

仮定:

cos (x) = u

3 u^{2} - 6 u = - 3

3 u^{2} - 6 u = - 3 : u = 1

3 u^{2} - 6 u = - 3

3

を左側に移動します

3 u^{2} - 6 u = - 3

両辺に

3

を足す

3 u^{2} - 6 u + 3 = - 3 + 3

簡素化

3 u^{2} - 6 u + 3 = 0

3 u^{2} - 6 u + 3 = 0

解くとthe二次式

3 u^{2} - 6 u + 3 = 0

二次Equationの公式:

次の場合:

a = 3, b = - 6, c = 3

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

(- 6)^{2} - 4 \cdot 3 \cdot 3 = 0

(- 6)^{2} - 4 \cdot 3 \cdot 3

指数の規則を適用する:

n

が偶数であれば

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

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 3 \cdot 3

数を乗じる:

4 \cdot 3 \cdot 3 = 36

= 6^{2} - 36

6^{2} = 36

= 36 - 36

数を引く:

36 - 36 = 0

= 0

u_{1, 2} = \frac{- ( - 6 ) \pm 0}{2 \cdot 3}

u = \frac{- ( - 6 )}{2 \cdot 3}

\frac{- ( - 6 )}{2 \cdot 3} = 1

\frac{- ( - 6 )}{2 \cdot 3}

規則を適用

- (- a) = a

= \frac{6}{2 \cdot 3}

数を乗じる:

2 \cdot 3 = 6

= \frac{6}{6}

規則を適用

\frac{a}{a} = 1

= 1

u = 1

二次equationの解:

u = 1

代用を戻す

u = cos (x)

cos (x) = 1

cos (x) = 1

cos (x) = 1, 0 \leq x \leq 2 π : x = 0, x = 2 π

cos (x) = 1, 0 \leq x \leq 2 π

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 = 0 + 2 πn

x = 0 + 2 πn

解く

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn

範囲の解答

0 \leq x \leq 2 π

x = 0, x = 2 π

すべての解を組み合わせる

x = 0, x = 2 π

グラフ

人気の例

csc(θ)+2=0,0<= θ<= 2pi

csc (θ) + 2 = 0, 0 \leq θ \leq 2 π

3sin^2(θ)-sin(θ)-4=0

3 sin^{2} (θ) - sin (θ) - 4 = 0

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

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

sqrt(3)tan(x)-2=-1

3 tan (x) - 2 = - 1

solvefor x,1-cos(x)=0

so l v e f or x, 1 - cos (x) = 0