What is the general solution for 4cos^2(2x)-4cos(2x)+1=0 ?

The general solution for 4cos^2(2x)-4cos(2x)+1=0 is x= pi/6+pin,x=(5pi)/6+pin

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4cos^2(2x)-4cos(2x)+1=0

4 cos^{2} (2 x) - 4 cos (2 x) + 1 = 0

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Solution steps

4 cos^{2} (2 x) - 4 cos (2 x) + 1 = 0

Solve by substitution

cos (2 x) = \frac{1}{2}

cos (2 x) = \frac{1}{2} : x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

cos (θ) = \frac{2}{13}

4 sin (x - π) + 2 = 0

\frac{1}{2} tan (2 x) = tan (x)

sin (x + π) + cos (x + π) = 0

2 sin^{2} (x) - 3 sin (x) = - 1, 0 \leq x \leq 2 π

What is the general solution for 4cos^2(2x)-4cos(2x)+1=0 ?
The general solution for 4cos^2(2x)-4cos(2x)+1=0 is x= pi/6+pin,x=(5pi)/6+pin