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受欢迎的三角函数 >

2sin(2θ)+sqrt(3)=0

解答

2 sin (2 θ) + 3 = 0

解答

θ = \frac{2 π}{3} + πn, θ = \frac{5 π}{6} + πn

+1

度数

θ = 12 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n

求解步骤

2 sin (2 θ) + 3 = 0

将

3

到右边

2 sin (2 θ) + 3 = 0

两边减去

3

2 sin (2 θ) + 3 - 3 = 0 - 3

化简

2 sin (2 θ) = - 3

2 sin (2 θ) = - 3

两边除以

2

2 sin (2 θ) = - 3

两边除以

2

\frac{2 sin ( 2 θ )}{2} = \frac{- 3}{2}

化简

sin (2 θ) = - \frac{3}{2}

sin (2 θ) = - \frac{3}{2}

sin (2 θ) = - \frac{3}{2}

的通解

sin (x)

周期表（周期为

2 πn

"）：

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

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

解

2 θ = \frac{4 π}{3} + 2 πn : θ = \frac{2 π}{3} + πn

2 θ = \frac{4 π}{3} + 2 πn

两边除以

2

2 θ = \frac{4 π}{3} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2} : \frac{2 π}{3} + πn

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

使用分式法则:

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

= \frac{4 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{4 π}{6}

约分:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

解

2 θ = \frac{5 π}{3} + 2 πn : θ = \frac{5 π}{6} + πn

2 θ = \frac{5 π}{3} + 2 πn

两边除以

2

2 θ = \frac{5 π}{3} + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

化简

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

数字相除:

\frac{2}{2} = 1

= θ

化简

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

使用分式法则:

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

= \frac{5 π}{3 \cdot 2}

数字相乘:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

数字相除:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

θ = \frac{2 π}{3} + πn, θ = \frac{5 π}{6} + πn

作图

流行的例子

sin^2(x)+4sin(x)+3=0

sin^{2} (x) + 4 sin (x) + 3 = 0

cos^2(x)-sin^2(x)=(sqrt(2))/2

cos^{2} (x) - sin^{2} (x) = \frac{2}{2}

4cos(2θ)+3cos(θ)+4=-3cos(θ)-1

4 cos (2 θ) + 3 cos (θ) + 4 = - 3 cos (θ) - 1

cos(θ-pi/4)-1=0

cos (θ - \frac{π}{4}) - 1 = 0

solvefor x,-sin(x)=0

so l v e f or x, - sin (x) = 0