What is the general solution for cos(2x)=0,x<= 2pi,0 ?

The general solution for cos(2x)=0,x<= 2pi,0 is

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cos(2x)=0,x<= 2pi,0

cos (2 x) = 0, x \leq 2 π, 0

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}, x = - \frac{7 π}{4}, x = - \frac{5 π}{4}, x = - \frac{11 π}{4}, x = - \frac{9 π}{4}, x = - \frac{15 π}{4}, x = - \frac{13 π}{4}

Solution steps

cos (2 x) = 0, x \leq 2 π, 0

General solutions for

cos (2 x) = 0

2 x = \frac{π}{2} + 2 πn, 2 x = \frac{3 π}{2} + 2 πn

Solve

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

Solve

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

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Solutions for the range

x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}, x = - \frac{3 π}{4}, x = - \frac{π}{4}, x = - \frac{7 π}{4}, x = - \frac{5 π}{4}, x = - \frac{11 π}{4}, x = - \frac{9 π}{4}, x = - \frac{15 π}{4}, x = - \frac{13 π}{4}

7 + csc (2 x) = 3 cot^{2} (2 x)

cos (θ) = cos (\frac{π}{13})

sin (2 x + \frac{π}{6}) = - 1

tan (x) = 2 π

2 sin^{2} (x) - 2 = 0

What is the general solution for cos(2x)=0,x<= 2pi,0 ?
The general solution for cos(2x)=0,x<= 2pi,0 is