Solve cos^2x * tan^2x - 1.5 = 0 for 0 to 360 degrees. Please help, Ive been stuck on this for a while

As I read this, I think there are no solutions.
(cos x)^2*(tan x)^2=(sin x)^2 since tan x=sinx/cosx.

Then you would have
(sin x)^2-1,5=0
(sinx )^2=1,5

sin x= +- sqrt(1.5) = +- 1.225 but -1 <= sin x <=1 for all x.

First, observe that:

cos²(x)tan²(x) = sin²(x)

and then arrange the equation as:

sin²(x) = 3/2

hence:

sin(x) = ±√(3)/2

First quadrant: x = 60°

Second quadrant: x = 120°

Third quadrant: x = 240°

Fourth quadrant: x = 300°

tan^2(x) = sin^2(x)/cos^2(x)

Therefore
cos^2(x) * tan^2(x) = sin^2(x)

sin^2(x) + cos^2(x) = 1
therefore

cos^2(x) * tan^2(x) - 1.5 = 0
can be changed to
sin^2 - 1 = 0.5
cos^2(x) = 0.5
cos(x) = sqrt(0.5)

x = pi/4 + k(pi/2)
where k is any integer

Ummm...
tan = sin/cos
so cos^2 * tan^2 = cos^2 * sin^2/cos^2 = sin^2
so sin^2(x) = 1.5 ==> |sin(x)| = √(1.5) > 1
which is extraneous, so there is no solution.

sin^2(x) = 3/2 => sin(x) = +/- √(3/2)

No soln since sin(x) only ranges from -1 to 1.