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人気のある三角関数 >

(sin(2x)+1)=0,0<= x<360

解

(sin (2 x) + 1) = 0, 0^{\circ} \leq x < 36 0^{\circ}

解

x = 13 5^{\circ}, x = 31 5^{\circ}

+1

ラジアン

x = \frac{3 π}{4}, x = \frac{7 π}{4}

解答ステップ

(sin (2 x) + 1) = 0, 0^{\circ} \leq x < 36 0^{\circ}

1

を右側に移動します

sin (2 x) + 1 = 0

両辺から

1

を引く

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

簡素化

sin (2 x) = - 1

sin (2 x) = - 1

以下の一般解

sin (2 x) = - 1

sin (x)

36 0^{\circ} n

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}

2 x = 27 0^{\circ} + 36 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

解く

2 x = 27 0^{\circ} + 36 0^{\circ} n : x = 13 5^{\circ} + 18 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

以下で両辺を割る

2

2 x = 27 0^{\circ} + 36 0^{\circ} n

以下で両辺を割る

2

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

簡素化

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

簡素化

\frac{2 x}{2} : x

\frac{2 x}{2}

数を割る:

\frac{2}{2} = 1

= x

簡素化

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

分数の規則を適用する:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{54 0 ^{\circ}}{2 \cdot 2}

数を乗じる:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

数を割る:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

範囲の解答

0 \leq x < 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

グラフ

人気の例

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

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

solvefor y,-1/2 cot(y)=(t^2)/2+C

so l v e f or y, - \frac{1}{2} cot (y) = \frac{t ^{2}}{2} + C

4cos(2t)+1=3,0, pi/2

4 cos (2 t) + 1 = 3, 0, \frac{π}{2}

solvefor A,cos(pi/3-A)=2sin(A)

so l v e f or A, cos (\frac{π}{3} - A) = 2 sin (A)

-sqrt(3)+3=2sqrt(6)cos(θ)

- 3 + 3 = 26 cos (θ)