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人気のある三角関数 >

2sin^2(t)-cos(t)=1

解

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

解

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

+1

度

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

解答ステップ

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

両辺から

1

を引く

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

三角関数の公式を使用して書き換える

- 1 - cos (t) + 2 sin^{2} (t)

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 (t) + 2 (1 - cos^{2} (t))

簡素化

- 1 - cos (t) + 2 (1 - cos^{2} (t)) : - 2 cos^{2} (t) - cos (t) + 1

- 1 - cos (t) + 2 (1 - cos^{2} (t))

拡張

2 (1 - cos^{2} (t)) : 2 - 2 cos^{2} (t)

2 (1 - cos^{2} (t))

分配法則を適用する:

a (b - c) = ab - a c

a = 2, b = 1, c = cos^{2} (t)

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

数を乗じる:

2 \cdot 1 = 2

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

= - 1 - cos (t) + 2 - 2 cos^{2} (t)

簡素化

- 1 - cos (t) + 2 - 2 cos^{2} (t) : - 2 cos^{2} (t) - cos (t) + 1

- 1 - cos (t) + 2 - 2 cos^{2} (t)

条件のようなグループ

= - cos (t) - 2 cos^{2} (t) - 1 + 2

数を足す/引く:

- 1 + 2 = 1

= - 2 cos^{2} (t) - cos (t) + 1

= - 2 cos^{2} (t) - cos (t) + 1

= - 2 cos^{2} (t) - cos (t) + 1

1 - cos (t) - 2 cos^{2} (t) = 0

置換で解く

1 - cos (t) - 2 cos^{2} (t) = 0

仮定:

cos (t) = u

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

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

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

標準的な形式で書く

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

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

解くとthe二次式

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

二次Equationの公式:

次の場合:

a = - 2, b = - 1, c = 1

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

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

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

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

規則を適用

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

指数の規則を適用する:

n

が偶数であれば

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

(- 1)^{2} = 1^{2}

= 1^{2}

規則を適用

1^{a} = 1

= 1

4 \cdot 2 \cdot 1 = 8

4 \cdot 2 \cdot 1

数を乗じる:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

数を足す:

1 + 8 = 9

= 9

数を因数に分解する:

9 = 3^{2}

= 3^{2}

累乗根の規則を適用する:

n a^{n} = a

3^{2} = 3

= 3

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

解を分離する

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

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

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

括弧を削除する:

(- a) = - a, - (- a) = a

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

数を足す:

1 + 3 = 4

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

数を乗じる:

2 \cdot 2 = 4

= \frac{4}{- 4}

分数の規則を適用する:

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

= - \frac{4}{4}

規則を適用

\frac{a}{a} = 1

= - 1

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

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

括弧を削除する:

(- a) = - a, - (- a) = a

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

数を引く:

1 - 3 = - 2

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

数を乗じる:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

分数の規則を適用する:

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

= \frac{2}{4}

共通因数を約分する:

2

= \frac{1}{2}

二次equationの解:

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

代用を戻す

u = cos (t)

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

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

cos (t) = - 1 : t = π + 2 πn

cos (t) = - 1

以下の一般解

cos (t) = - 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}

t = π + 2 πn

t = π + 2 πn

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

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

以下の一般解

cos (t) = \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}

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

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

すべての解を組み合わせる

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

グラフ

人気の例

sin(θ)= 8/17 ,cos(θ)

sin (θ) = \frac{8}{17}, cos (θ)

solvefor θ,3cos(θ-55)=-1.56

so l v e f or θ, 3 cos (θ - 5 5^{\circ}) = - 1.56

-6sin(c)-12=2sin(c)-4

- 6 sin (c) - 12 = 2 sin (c) - 4

tan(2x+10)=cot(3x-15)

tan (2 x + 10) = cot (3 x - 15)

tan (θ) = \frac{- 4}{3}