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人気のある三角関数 >

(cos(t)-4)(2sin^2(t)-1)=0

解

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0

解

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

+1

度

t = 4 5^{\circ} + 36 0^{\circ} n, t = 13 5^{\circ} + 36 0^{\circ} n, t = 22 5^{\circ} + 36 0^{\circ} n, t = 31 5^{\circ} + 36 0^{\circ} n

解答ステップ

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0

各部分を別個に解く

cos (t) - 4 = 0 or 2 sin^{2} (t) - 1 = 0

cos (t) - 4 = 0 :

解なし

cos (t) - 4 = 0

4

を右側に移動します

cos (t) - 4 = 0

両辺に

4

を足す

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

簡素化

cos (t) = 4

cos (t) = 4

- 1 \leq cos (x) \leq 1

解なし

2 sin^{2} (t) - 1 = 0 : t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

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

置換で解く

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

仮定:

sin (t) = u

2 u^{2} - 1 = 0

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

2 u^{2} - 1 = 0

1

を右側に移動します

2 u^{2} - 1 = 0

両辺に

1

を足す

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

簡素化

2 u^{2} = 1

2 u^{2} = 1

以下で両辺を割る

2

2 u^{2} = 1

以下で両辺を割る

2

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

簡素化

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

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

x^{2} = f (a)

の場合, 解は

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

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

代用を戻す

u = sin (t)

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

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

sin (t) = \frac{1}{2} : t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn

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

以下の一般解

sin (t) = \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}

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn

sin (t) = - \frac{1}{2} : t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

sin (t) = - \frac{1}{2}

以下の一般解

sin (t) = - \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}

t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

すべての解を組み合わせる

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

すべての解を組み合わせる

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

グラフ

人気の例

sin (7 5^{\circ}) = \frac{x}{9}

solvefor i,2cos^3(x)+sin(x)+1=2sin^2(x)

so l v e f or i, 2 cos^{3} (x) + sin (x) + 1 = 2 sin^{2} (x)

solvefor t,tan(x)=csc^2(x)-2

so l v e f or t, tan (x) = csc^{2} (x) - 2

solvefor y,xy+2y=sin(x)

so l v e f or y, x y + 2 y = sin (x)

cos(15)= a/(55)

cos (1 5^{\circ}) = \frac{a}{55}