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人気のある三角関数 >

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

解

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

解

x = 2 πn, x = π + 2 πn

+1

度

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

解答ステップ

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

三角関数の公式を使用して書き換える

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

2倍角の公式を使用:

cos (2 x) = 1 - 2 sin^{2} (x)

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

簡素化

- 1 + (1 - 2 sin^{2} (x))^{2} - 3 sin^{2} (x) : 4 sin^{4} (x) - 7 sin^{2} (x)

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

(1 - 2 sin^{2} (x))^{2} : 1 - 4 sin^{2} (x) + 4 sin^{4} (x)

完全平方式を適用する:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = 1, b = 2 sin^{2} (x)

= 1^{2} - 2 \cdot 1 \cdot 2 sin^{2} (x) + (2 sin^{2} (x))^{2}

簡素化

1^{2} - 2 \cdot 1 \cdot 2 sin^{2} (x) + (2 sin^{2} (x))^{2} : 1 - 4 sin^{2} (x) + 4 sin^{4} (x)

1^{2} - 2 \cdot 1 \cdot 2 sin^{2} (x) + (2 sin^{2} (x))^{2}

規則を適用

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot 2 sin^{2} (x) + (2 sin^{2} (x))^{2}

2 \cdot 1 \cdot 2 sin^{2} (x) = 4 sin^{2} (x)

2 \cdot 1 \cdot 2 sin^{2} (x)

数を乗じる:

2 \cdot 1 \cdot 2 = 4

= 4 sin^{2} (x)

(2 sin^{2} (x))^{2} = 4 sin^{4} (x)

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

指数の規則を適用する:

(a \cdot b)^{n} = a^{n} b^{n}

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

(sin^{2} (x))^{2} : sin^{4} (x)

指数の規則を適用する:

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

= sin^{2 \cdot 2} (x)

数を乗じる:

2 \cdot 2 = 4

= sin^{4} (x)

= 2^{2} sin^{4} (x)

2^{2} = 4

= 4 sin^{4} (x)

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

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

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

簡素化

- 1 + 1 - 4 sin^{2} (x) + 4 sin^{4} (x) - 3 sin^{2} (x) : 4 sin^{4} (x) - 7 sin^{2} (x)

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

条件のようなグループ

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

類似した元を足す:

- 4 sin^{2} (x) - 3 sin^{2} (x) = - 7 sin^{2} (x)

= - 7 sin^{2} (x) + 4 sin^{4} (x) - 1 + 1

- 1 + 1 = 0

= 4 sin^{4} (x) - 7 sin^{2} (x)

= 4 sin^{4} (x) - 7 sin^{2} (x)

= 4 sin^{4} (x) - 7 sin^{2} (x)

4 sin^{4} (x) - 7 sin^{2} (x) = 0

置換で解く

4 sin^{4} (x) - 7 sin^{2} (x) = 0

仮定:

sin (x) = u

4 u^{4} - 7 u^{2} = 0

4 u^{4} - 7 u^{2} = 0 : u = \frac{7}{2}, u = - \frac{7}{2}, u = 0

4 u^{4} - 7 u^{2} = 0

equationを

v = u^{2}

と以下で書き換える:

v^{2} = u^{4}

4 v^{2} - 7 v = 0

解く

4 v^{2} - 7 v = 0 : v = \frac{7}{4}, v = 0

4 v^{2} - 7 v = 0

解くとthe二次式

4 v^{2} - 7 v = 0

二次Equationの公式:

次の場合:

a = 4, b = - 7, c = 0

v_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

v_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 7)^{2} - 4 \cdot 4 \cdot 0 = 7

(- 7)^{2} - 4 \cdot 4 \cdot 0

指数の規則を適用する:

n

が偶数であれば

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

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 4 \cdot 0

規則を適用

0 \cdot a = 0

= 7^{2} - 0

7^{2} - 0 = 7^{2}

= 7^{2}

累乗根の規則を適用する:

n a^{n} = a,

, 以下を想定

a \geq 0

= 7

v_{1, 2} = \frac{- ( - 7 ) \pm 7}{2 \cdot 4}

解を分離する

v_{1} = \frac{- ( - 7 ) + 7}{2 \cdot 4}, v_{2} = \frac{- ( - 7 ) - 7}{2 \cdot 4}

v = \frac{- ( - 7 ) + 7}{2 \cdot 4} : \frac{7}{4}

\frac{- ( - 7 ) + 7}{2 \cdot 4}

規則を適用

- (- a) = a

= \frac{7 + 7}{2 \cdot 4}

数を足す:

7 + 7 = 14

= \frac{14}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{14}{8}

共通因数を約分する:

2

= \frac{7}{4}

v = \frac{- ( - 7 ) - 7}{2 \cdot 4} : 0

\frac{- ( - 7 ) - 7}{2 \cdot 4}

規則を適用

- (- a) = a

= \frac{7 - 7}{2 \cdot 4}

数を引く:

7 - 7 = 0

= \frac{0}{2 \cdot 4}

数を乗じる:

2 \cdot 4 = 8

= \frac{0}{8}

規則を適用

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

= 0

二次equationの解:

v = \frac{7}{4}, v = 0

v = \frac{7}{4}, v = 0

再び

v = u^{2}

に置き換えて以下を解く:

u

解く

u^{2} = \frac{7}{4} : u = \frac{7}{2}, u = - \frac{7}{2}

u^{2} = \frac{7}{4}

x^{2} = f (a)

の場合, 解は

x = f (a), - f (a)

u = \frac{7}{4}, u = - \frac{7}{4}

\frac{7}{4} = \frac{7}{2}

\frac{7}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{7}{2}

- \frac{7}{4} = - \frac{7}{2}

- \frac{7}{4}

簡素化

\frac{7}{4} : \frac{7}{2}

\frac{7}{4}

累乗根の規則を適用する:

n \frac{a}{b} = \frac{n a}{n b},

, 以下を想定

a \geq 0, b \geq 0

= \frac{7}{4}

4 = 2

4

数を因数に分解する:

4 = 2^{2}

= 2^{2}

累乗根の規則を適用する:

n a^{n} = a

2^{2} = 2

= 2

= \frac{7}{2}

= - \frac{7}{2}

u = \frac{7}{2}, u = - \frac{7}{2}

解く

u^{2} = 0 : u = 0

u^{2} = 0

規則を適用

x^{n} = 0 \Rightarrow x = 0

u = 0

解答は

u = \frac{7}{2}, u = - \frac{7}{2}, u = 0

代用を戻す

u = sin (x)

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}, sin (x) = 0

sin (x) = \frac{7}{2}, sin (x) = - \frac{7}{2}, sin (x) = 0

sin (x) = \frac{7}{2} :

解なし

sin (x) = \frac{7}{2}

- 1 \leq sin (x) \leq 1

解なし

sin (x) = - \frac{7}{2} :

解なし

sin (x) = - \frac{7}{2}

- 1 \leq sin (x) \leq 1

解なし

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

以下の一般解

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 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, x = π + 2 πn

すべての解を組み合わせる

x = 2 πn, x = π + 2 πn

グラフ

人気の例

sin (- t) = \frac{3}{10}

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

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

solvefor c,e=r(sec(c/2)-1)

so l v e f or c, e = r (sec (\frac{c}{2}) - 1)

arctan(x/3)+arctan(x/2)=arctan(x)

arctan (\frac{x}{3}) + arctan (\frac{x}{2}) = arctan (x)

solvefor w,y=arctan(1+4w)

so l v e f or w, y = arctan (1 + 4 w)