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人気のある三角関数 >

8cos^2(θ)-3cos(θ)=0,0<= θ<= 180

解

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

解

θ = 1.18639 \dots, θ = 9 0^{\circ}

+1

ラジアン

θ = 1.18639 \dots, θ = \frac{π}{2}

解答ステップ

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

置換で解く

8 cos^{2} (θ) - 3 cos (θ) = 0

仮定:

cos (θ) = u

8 u^{2} - 3 u = 0

8 u^{2} - 3 u = 0 : u = \frac{3}{8}, u = 0

8 u^{2} - 3 u = 0

解くとthe二次式

8 u^{2} - 3 u = 0

二次Equationの公式:

次の場合:

a = 8, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 8 \cdot 0}{2 \cdot 8}

(- 3)^{2} - 4 \cdot 8 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 8 \cdot 0

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 8 \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 8}

解を分離する

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 8}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 8}

u = \frac{- ( - 3 ) + 3}{2 \cdot 8} : \frac{3}{8}

\frac{- ( - 3 ) + 3}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{3 + 3}{2 \cdot 8}

数を足す:

3 + 3 = 6

= \frac{6}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{6}{16}

共通因数を約分する:

2

= \frac{3}{8}

u = \frac{- ( - 3 ) - 3}{2 \cdot 8} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 8}

規則を適用

- (- a) = a

= \frac{3 - 3}{2 \cdot 8}

数を引く:

3 - 3 = 0

= \frac{0}{2 \cdot 8}

数を乗じる:

2 \cdot 8 = 16

= \frac{0}{16}

規則を適用

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

= 0

二次equationの解:

u = \frac{3}{8}, u = 0

代用を戻す

u = cos (θ)

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, cos (θ) = 0

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ} : θ = arccos (\frac{3}{8})

cos (θ) = \frac{3}{8}, 0 \leq θ \leq 18 0^{\circ}

三角関数の逆数プロパティを適用する

cos (θ) = \frac{3}{8}

以下の一般解

cos (θ) = \frac{3}{8}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

θ = arccos (\frac{3}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{8}) + 36 0^{\circ} n

θ = arccos (\frac{3}{8}) + 36 0^{\circ} n, θ = 36 0^{\circ} - arccos (\frac{3}{8}) + 36 0^{\circ} n

範囲の解答

0 \leq θ \leq 18 0^{\circ}

θ = arccos (\frac{3}{8})

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ} : θ = 9 0^{\circ}

cos (θ) = 0, 0 \leq θ \leq 18 0^{\circ}

以下の一般解

cos (θ) = 0

cos (x)

36 0^{\circ} n

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}

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

範囲の解答

0 \leq θ \leq 18 0^{\circ}

θ = 9 0^{\circ}

すべての解を組み合わせる

θ = arccos (\frac{3}{8}), θ = 9 0^{\circ}

10進法形式で解を証明する

θ = 1.18639 \dots, θ = 9 0^{\circ}

グラフ

人気の例

8 sin (x) - 7 = 0

sec (t) = 2

tan(x+2)=cot(4x+28)

tan (x + 2) = cot (4 x + 28)

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

2 tan (x) + 5 = 3