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人気のある三角関数 >

5cos^2(x)+3cos(x)=0

解

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

解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

+1

度

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 126.86989 \dots^{\circ} + 36 0^{\circ} n, x = - 126.86989 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

置換で解く

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

仮定:

cos (x) = u

5 u^{2} + 3 u = 0

5 u^{2} + 3 u = 0 : u = 0, u = - \frac{3}{5}

5 u^{2} + 3 u = 0

解くとthe二次式

5 u^{2} + 3 u = 0

二次Equationの公式:

次の場合:

a = 5, b = 3, c = 0

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

3^{2} - 4 \cdot 5 \cdot 0 = 3

3^{2} - 4 \cdot 5 \cdot 0

規則を適用

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 3

u_{1, 2} = \frac{- 3 \pm 3}{2 \cdot 5}

解を分離する

u_{1} = \frac{- 3 + 3}{2 \cdot 5}, u_{2} = \frac{- 3 - 3}{2 \cdot 5}

u = \frac{- 3 + 3}{2 \cdot 5} : 0

\frac{- 3 + 3}{2 \cdot 5}

数を足す/引く:

- 3 + 3 = 0

= \frac{0}{2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{0}{10}

規則を適用

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 3 - 3}{2 \cdot 5} : - \frac{3}{5}

\frac{- 3 - 3}{2 \cdot 5}

数を引く:

- 3 - 3 = - 6

= \frac{- 6}{2 \cdot 5}

数を乗じる:

2 \cdot 5 = 10

= \frac{- 6}{10}

分数の規則を適用する:

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

= - \frac{6}{10}

共通因数を約分する:

2

= - \frac{3}{5}

二次equationの解:

u = 0, u = - \frac{3}{5}

代用を戻す

u = cos (x)

cos (x) = 0, cos (x) = - \frac{3}{5}

cos (x) = 0, cos (x) = - \frac{3}{5}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

以下の一般解

cos (x) = 0

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

cos (x) = - \frac{3}{5} : x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

cos (x) = - \frac{3}{5}

三角関数の逆数プロパティを適用する

cos (x) = - \frac{3}{5}

以下の一般解

cos (x) = - \frac{3}{5}

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

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

すべての解を組み合わせる

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (- \frac{3}{5}) + 2 πn, x = - arccos (- \frac{3}{5}) + 2 πn

10進法形式で解を証明する

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 2.21429 \dots + 2 πn, x = - 2.21429 \dots + 2 πn

グラフ

人気の例

-2/(sqrt(3))cos(3θ)=1

- \frac{2}{3} cos (3 θ) = 1

solvefor x,y=sin(xy)

so l v e f or x, y = sin (x y)

tan(t)= 1/(sqrt(3))

tan (t) = \frac{1}{3}

4sin^2(x)+3sin(x)-1=0

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

cos (x) = \frac{π}{2}