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受欢迎的三角函数 >

2cos(2x)-2sin(2x)=0

解答

2 cos (2 x) - 2 sin (2 x) = 0

解答

x = \frac{π}{8} + \frac{πn}{2}

+1

度数

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

求解步骤

2 cos (2 x) - 2 sin (2 x) = 0

使用三角恒等式改写

2 cos (2 x) - 2 sin (2 x) = 0

在两边除以

cos (2 x), cos (2 x) \neq = 0

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

化简

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

使用基本三角恒等式:

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

2 - 2 tan (2 x) = 0

2 - 2 tan (2 x) = 0

将

2

到右边

2 - 2 tan (2 x) = 0

两边减去

2

2 - 2 tan (2 x) - 2 = 0 - 2

化简

- 2 tan (2 x) = - 2

- 2 tan (2 x) = - 2

两边除以

- 2

- 2 tan (2 x) = - 2

两边除以

- 2

\frac{- 2 tan ( 2 x )}{- 2} = \frac{- 2}{- 2}

化简

tan (2 x) = 1

tan (2 x) = 1

tan (2 x) = 1

的通解

tan (x)

周期表（周期为

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

解

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

2 x = \frac{π}{4} + πn

两边除以

2

2 x = \frac{π}{4} + πn

两边除以

2

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

化简

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

化简

\frac{2 x}{2} : x

\frac{2 x}{2}

数字相除:

\frac{2}{2} = 1

= x

化简

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

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

使用分式法则:

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

= \frac{π}{4 \cdot 2}

数字相乘:

4 \cdot 2 = 8

= \frac{π}{8}

= \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

流行的例子

2cos(x/2)-sqrt(2)=0,0<= x<2pi

2 cos (\frac{x}{2}) - 2 = 0, 0 \leq x < 2 π

cos(x/4)=-(sqrt(2))/2

cos (\frac{x}{4}) = - \frac{2}{2}

tan (x) = \frac{12}{20}

15 cos (x) = 0

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

sin (\frac{π}{2} - x) = cos^{2} (x)