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人気のある三角関数 >

sin(x)cos(x^2)=sin(x),0<= x<= 2pi

解

sin (x) cos (x^{2}) = sin (x), 0 \leq x \leq 2 π

解

x = 0, x = π, x = 2 π, x = 2 π, x = 2 π, x = 6 π, x = 22 π

+1

度

x = 0^{\circ}, x = 18 0^{\circ}, x = 36 0^{\circ}, x = 143.61922 \dots^{\circ}, x = 203.10825 \dots^{\circ}, x = 248.75578 \dots^{\circ}, x = 287.23844 \dots^{\circ}

解答ステップ

sin (x) cos (x^{2}) = sin (x), 0 \leq x \leq 2 π

両辺から

sin (x)

を引く

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

因数

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

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

共通項をくくり出す

sin (x)

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

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

各部分を別個に解く

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

sin (x) = 0, 0 \leq x \leq 2 π : x = 0, x = π, x = 2 π

sin (x) = 0, 0 \leq x \leq 2 π

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

範囲の解答

0 \leq x \leq 2 π

x = 0, x = π, x = 2 π

cos (x^{2}) - 1 = 0, 0 \leq x \leq 2 π : x = 0, x = 0, x = 2 π, x = 2 π, x = 6 π, x = 22 π

cos (x^{2}) - 1 = 0, 0 \leq x \leq 2 π

1

を右側に移動します

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

両辺に

1

を足す

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

簡素化

cos (x^{2}) = 1

cos (x^{2}) = 1

以下の一般解

cos (x^{2}) = 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^{2} = 0 + 2 πn

x^{2} = 0 + 2 πn

解く

x^{2} = 0 + 2 πn : x = 2 πn, x = - 2 πn

x^{2} = 0 + 2 πn

0 + 2 πn = 2 πn

x^{2} = 2 πn

x^{2} = f (a)

の場合, 解は

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

x = 2 πn, x = - 2 πn

簡素化

2 πn : 2 πn

2 πn

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= 2 πn

簡素化

- 2 πn : - 2 πn

- 2 πn

累乗根の規則を適用する:

n ab = n a n b,

, 以下を想定

a \geq 0, b \geq 0

= - 2 πn

x = 2 πn, x = - 2 πn

x = 2 πn, x = - 2 πn

範囲の解答

0 \leq x \leq 2 π

x = 0, x = 0, x = 2 π, x = 2 π, x = 6 π, x = 22 π

すべての解を組み合わせる

x = 0, x = π, x = 2 π, x = 2 π, x = 2 π, x = 6 π, x = 22 π

グラフ

人気の例

solvefor θ,(cos(θ))/(sin(θ))=-pi/6

so l v e f or θ, \frac{cos ( θ )}{sin ( θ )} = - \frac{π}{6}

tan(θ)=(10)/(-5.32)

tan (θ) = \frac{10}{- 5.32}

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

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

sec(t)= 1/((sqrt(2))/2)

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

-(cos(x)-sin(x))/(1+sin(2x))=0

- \frac{cos ( x ) - sin ( x )}{1 + sin ( 2 x )} = 0