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人気のある三角関数 >

1+2cos(x)+cos^2(x)=3sin^2(x)

解

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

解

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

+1

度

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

解答ステップ

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

両辺から

3 sin^{2} (x)

を引く

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

三角関数の公式を使用して書き換える

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

ピタゴラスの公式を使用する:

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

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

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

簡素化

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

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

拡張

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

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

分配法則を適用する:

a (b - c) = ab - a c

a = - 3, b = 1, c = cos^{2} (x)

= - 3 \cdot 1 - (- 3) cos^{2} (x)

マイナス・プラスの規則を適用する

- (- a) = a

= - 3 \cdot 1 + 3 cos^{2} (x)

数を乗じる:

3 \cdot 1 = 3

= - 3 + 3 cos^{2} (x)

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

簡素化

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

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

条件のようなグループ

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

類似した元を足す:

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

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

数を足す/引く:

1 - 3 = - 2

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

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

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

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

置換で解く

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

仮定:

cos (x) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = 4, b = 2, c = - 2

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

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

2^{2} - 4 \cdot 4 (- 2) = 6

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

規則を適用

- (- a) = a

= 2^{2} + 4 \cdot 4 \cdot 2

数を乗じる:

4 \cdot 4 \cdot 2 = 32

= 2^{2} + 32

2^{2} = 4

= 4 + 32

数を足す:

4 + 32 = 36

= 36

数を因数に分解する:

36 = 6^{2}

= 6^{2}

累乗根の規則を適用する:

n a^{n} = a

6^{2} = 6

= 6

u_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 4}

解を分離する

u_{1} = \frac{- 2 + 6}{2 \cdot 4}, u_{2} = \frac{- 2 - 6}{2 \cdot 4}

u = \frac{- 2 + 6}{2 \cdot 4} : \frac{1}{2}

\frac{- 2 + 6}{2 \cdot 4}

数を足す/引く:

- 2 + 6 = 4

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

数を乗じる:

2 \cdot 4 = 8

= \frac{4}{8}

共通因数を約分する:

4

= \frac{1}{2}

u = \frac{- 2 - 6}{2 \cdot 4} : - 1

\frac{- 2 - 6}{2 \cdot 4}

数を引く:

- 2 - 6 = - 8

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

数を乗じる:

2 \cdot 4 = 8

= \frac{- 8}{8}

分数の規則を適用する:

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

= - \frac{8}{8}

規則を適用

\frac{a}{a} = 1

= - 1

二次equationの解:

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

代用を戻す

u = cos (x)

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

以下の一般解

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

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

以下の一般解

cos (x) = - 1

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 = π + 2 πn

x = π + 2 πn

すべての解を組み合わせる

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn, x = π + 2 πn

グラフ

人気の例

solvefor L,sin(kL)=0

so l v e f or L, sin (k L) = 0

cos(x)=(9.8)/((0.1)(6.28)^2)

cos (x) = \frac{9.8}{( 0.1 ) ( 6.28 ) ^{2}}

-4sin(t)+5cos(t)=0

- 4 sin (t) + 5 cos (t) = 0

solvefor θ,r=2sin(θ)

so l v e f or θ, r = 2 sin (θ)

csc (3 θ) = - 1