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受欢迎的三角函数 >

solvefor x,sin(2x)*cos(2x)=0.5

解答

求解

x, sin (2 x) \cdot cos (2 x) = 0.5

解答

x = \frac{π}{8} + \frac{πn}{2}

+1

度数

x = 22. 5^{\circ} + 9 0^{\circ} n

求解步骤

sin (2 x) cos (2 x) = 0.5

使用三角恒等式改写

sin (2 x) cos (2 x)

使用倍角公式:

2 sin (x) cos (x) = sin (2 x)

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

= \frac{sin ( 2 \cdot 2 x )}{2}

\frac{sin ( 2 \cdot 2 x )}{2} = 0.5

在两边乘以

2

\frac{sin ( 2 \cdot 2 x )}{2} = 0.5

在两边乘以

2

\frac{2 sin ( 2 \cdot 2 x )}{2} = 0.5 \cdot 2

化简

sin (2 \cdot 2 x) = 1

sin (2 \cdot 2 x) = 1

sin (2 \cdot 2 x) = 1

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 \cdot 2 x = \frac{π}{2} + 2 πn

2 \cdot 2 x = \frac{π}{2} + 2 πn

解

2 \cdot 2 x = \frac{π}{2} + 2 πn : x = \frac{π}{8} + \frac{πn}{2}

2 \cdot 2 x = \frac{π}{2} + 2 πn

两边除以

4

2 \cdot 2 x = \frac{π}{2} + 2 πn

两边除以

4

\frac{2 \cdot 2 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

化简

\frac{2 \cdot 2 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4}

化简

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

数字相乘:

2 \cdot 2 = 4

= \frac{4 x}{4}

数字相除:

\frac{4}{4} = 1

= x

化简

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} : \frac{π}{8} + \frac{πn}{2}

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

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

使用分式法则:

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

= \frac{π}{2 \cdot 4}

数字相乘:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

约分:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

作图

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