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人気のある三角関数 >

4cos^2(x)=sin^2(x)+3

解

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

解

x = 0.46364 \dots + 2 πn, x = π - 0.46364 \dots + 2 πn, x = - 0.46364 \dots + 2 πn, x = π + 0.46364 \dots + 2 πn

+1

度

x = 26.56505 \dots^{\circ} + 36 0^{\circ} n, x = 153.43494 \dots^{\circ} + 36 0^{\circ} n, x = - 26.56505 \dots^{\circ} + 36 0^{\circ} n, x = 206.56505 \dots^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

sin^{2} (x) + 3

を引く

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

- 3 - sin^{2} (x) + 4 (1 - sin^{2} (x)) : - 5 sin^{2} (x) + 1

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

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

数を乗じる:

4 \cdot 1 = 4

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

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

簡素化

- 3 - sin^{2} (x) + 4 - 4 sin^{2} (x) : - 5 sin^{2} (x) + 1

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

条件のようなグループ

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

類似した元を足す:

- sin^{2} (x) - 4 sin^{2} (x) = - 5 sin^{2} (x)

= - 5 sin^{2} (x) - 3 + 4

数を足す/引く:

- 3 + 4 = 1

= - 5 sin^{2} (x) + 1

= - 5 sin^{2} (x) + 1

= - 5 sin^{2} (x) + 1

1 - 5 sin^{2} (x) = 0

置換で解く

1 - 5 sin^{2} (x) = 0

仮定:

sin (x) = u

1 - 5 u^{2} = 0

1 - 5 u^{2} = 0 : u = \frac{1}{5}, u = - \frac{1}{5}

1 - 5 u^{2} = 0

1

を右側に移動します

1 - 5 u^{2} = 0

両辺から

1

を引く

1 - 5 u^{2} - 1 = 0 - 1

簡素化

- 5 u^{2} = - 1

- 5 u^{2} = - 1

以下で両辺を割る

- 5

- 5 u^{2} = - 1

以下で両辺を割る

- 5

\frac{- 5 u ^{2}}{- 5} = \frac{- 1}{- 5}

簡素化

u^{2} = \frac{1}{5}

u^{2} = \frac{1}{5}

x^{2} = f (a)

の場合, 解は

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

u = \frac{1}{5}, u = - \frac{1}{5}

代用を戻す

u = sin (x)

sin (x) = \frac{1}{5}, sin (x) = - \frac{1}{5}

sin (x) = \frac{1}{5}, sin (x) = - \frac{1}{5}

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

sin (x) = \frac{1}{5}

三角関数の逆数プロパティを適用する

sin (x) = \frac{1}{5}

以下の一般解

sin (x) = \frac{1}{5}

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

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

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

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

sin (x) = - \frac{1}{5}

三角関数の逆数プロパティを適用する

sin (x) = - \frac{1}{5}

以下の一般解

sin (x) = - \frac{1}{5}

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

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

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

すべての解を組み合わせる

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

10進法形式で解を証明する

x = 0.46364 \dots + 2 πn, x = π - 0.46364 \dots + 2 πn, x = - 0.46364 \dots + 2 πn, x = π + 0.46364 \dots + 2 πn

グラフ

人気の例

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

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

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

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

cos(2x)+cos(2x)+1=0

cos (2 x) + cos (2 x) + 1 = 0

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

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

arccos(x)-arcsin(x)=arcsin(1-x)

arccos (x) - arcsin (x) = arcsin (1 - x)