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人気のある三角関数 >

8^{sin^2(x)}=4^{sin(x)-1/8}

解

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

解

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.16744 \dots + 2 πn, x = π - 0.16744 \dots + 2 πn

+1

度

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 9.59406 \dots^{\circ} + 36 0^{\circ} n, x = 170.40593 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

置換で解く

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

仮定:

sin (x) = u

8^{u^{2}} = 4^{u - \frac{1}{8}}

8^{u^{2}} = 4^{u - \frac{1}{8}} : u = \frac{1}{2}, u = \frac{1}{6}

8^{u^{2}} = 4^{u - \frac{1}{8}}

指数の規則を適用する

8^{u^{2}} = 4^{u - \frac{1}{8}}

底に変換する

2 : 2^{3 u^{2}} = 2^{2 (u - \frac{1}{8})}

4

を底に変換する

2

4 = 2^{2}

8^{u^{2}} = (2^{2})^{u - \frac{1}{8}}

8

を底に変換する

2

8 = 2^{3}

(2^{3})^{u^{2}} = (2^{2})^{u - \frac{1}{8}}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

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

2^{3 u^{2}} = (2^{2})^{u - \frac{1}{8}}

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

(2^{2})^{u - \frac{1}{8}} = 2^{2 (u - \frac{1}{8})}

2^{3 u^{2}} = 2^{2 (u - \frac{1}{8})}

2^{3 u^{2}} = 2^{2 (u - \frac{1}{8})}

a^{f (x)} = a^{g (x)}

ならば,

f (x) = g (x)

3 u^{2} = 2 (u - \frac{1}{8})

3 u^{2} = 2 (u - \frac{1}{8})

解く

3 u^{2} = 2 (u - \frac{1}{8}) : u = \frac{1}{2}, u = \frac{1}{6}

3 u^{2} = 2 (u - \frac{1}{8})

拡張

2 (u - \frac{1}{8}) : 2 u - \frac{1}{4}

2 (u - \frac{1}{8})

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = u, c = \frac{1}{8}

= 2 u - 2 \cdot \frac{1}{8}

2 \cdot \frac{1}{8} = \frac{1}{4}

2 \cdot \frac{1}{8}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{8}

数を乗じる:

1 \cdot 2 = 2

= \frac{2}{8}

共通因数を約分する:

2

= \frac{1}{4}

= 2 u - \frac{1}{4}

3 u^{2} = 2 u - \frac{1}{4}

\frac{1}{4}

を左側に移動します

3 u^{2} = 2 u - \frac{1}{4}

両辺に

\frac{1}{4}

を足す

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

簡素化

3 u^{2} + \frac{1}{4} = 2 u

3 u^{2} + \frac{1}{4} = 2 u

2 u

を左側に移動します

3 u^{2} + \frac{1}{4} = 2 u

両辺から

2 u

を引く

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

簡素化

3 u^{2} + \frac{1}{4} - 2 u = 0

3 u^{2} + \frac{1}{4} - 2 u = 0

標準的な形式で書く

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

3 u^{2} - 2 u + \frac{1}{4} = 0

解くとthe二次式

3 u^{2} - 2 u + \frac{1}{4} = 0

二次Equationの公式:

次の場合:

a = 3, b = - 2, c = \frac{1}{4}

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

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

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

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

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

(- 2)^{2}

指数の規則を適用する:

n

が偶数であれば

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

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

= 2^{2}

4 \cdot 3 \cdot \frac{1}{4} = 3

4 \cdot 3 \cdot \frac{1}{4}

分数を乗じる:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 4 \cdot 3}{4}

共通因数を約分する:

4

= 1 \cdot 3

数を乗じる:

1 \cdot 3 = 3

= 3

= 2^{2} - 3

2^{2} = 4

= 4 - 3

数を引く:

4 - 3 = 1

= 1

規則を適用

1 = 1

= 1

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

解を分離する

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

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

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

規則を適用

- (- a) = a

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

数を足す:

2 + 1 = 3

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

数を乗じる:

2 \cdot 3 = 6

= \frac{3}{6}

共通因数を約分する:

3

= \frac{1}{2}

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

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

規則を適用

- (- a) = a

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

数を引く:

2 - 1 = 1

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

数を乗じる:

2 \cdot 3 = 6

= \frac{1}{6}

二次equationの解:

u = \frac{1}{2}, u = \frac{1}{6}

u = \frac{1}{2}, u = \frac{1}{6}

代用を戻す

u = sin (x)

sin (x) = \frac{1}{2}, sin (x) = \frac{1}{6}

sin (x) = \frac{1}{2}, sin (x) = \frac{1}{6}

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

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

sin (x) = \frac{1}{6}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1}{6}

以下の一般解

sin (x) = \frac{1}{6}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

x = arcsin (\frac{1}{6}) + 2 πn, x = π - arcsin (\frac{1}{6}) + 2 πn

すべての解を組み合わせる

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

10進法形式で解を証明する

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = 0.16744 \dots + 2 πn, x = π - 0.16744 \dots + 2 πn

グラフ

人気の例

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

(sin(73))/(34)=(sin(d))/(29)

\frac{sin ( 7 3 ^{\circ} )}{34} = \frac{sin ( d )}{29}

solvefor θ,(90^2)/(250(32.2))=tan(θ)

so l v e f or θ, \frac{9 0 ^{2}}{250 ( 32.2 )} = tan (θ)

sin(θ)=2sin(θ)cos(θ)

sin (θ) = 2 sin (θ) cos (θ)

2 cos^{2} (4 x + 1) = 0