Solve 4 cos2x-3 = 0 for all real values of x.

4 cos² x - 3 = 0

4 cos² x = 3

cos² x = 3/4

cos x = ±(√3)/2

Fixing the squared cosine doesn't discriminate among quadrants.  There's one in every quadrant

cos x = ± cos(π/6)

Let's do plus first.  In general, cos x = cos a has solutions x = ±a + 2πk integer k

cos x = cos(π/6)

x = ±π/6 + 2πk

Minus next.

cos x = -cos(π/6)

cos x = cos(π - π/6)

cos x = cos(5π/6)

x = ±5π/6 + 2πk

We'll write all our solutions as

x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.

the awnser is visionary