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受欢迎的三角函数 >

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

解答

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

解答

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

+1

弧度

x = \frac{3 π}{4}, x = \frac{7 π}{4}

求解步骤

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

将

1

到右边

sin (2 x) + 1 = 0

两边减去

1

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

化简

sin (2 x) = - 1

sin (2 x) = - 1

sin (2 x) = - 1

的通解

sin (x)

周期表（周期为

36 0^{\circ} n

"）：

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} 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

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

两边除以

2

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

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

\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}

使用分式法则:

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

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

数字相乘:

2 \cdot 2 = 4

= 13 5^{\circ}

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

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

数字相除:

\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

在

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

范围内的解

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

作图

流行的例子

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)