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

The general solution for 2cos^2(2x)-1=0 is x= pi/8+pin,x=pi-pi/8+pin,x=(3pi)/8+pin,x=-(3pi)/8+pin

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

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

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

Solution steps

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

Solve by substitution

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

cos (2 x) = \frac{1}{2} : x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn

cos (2 x) = - \frac{1}{2} : x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

Combine all the solutions

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

5 cos (x) cot (x) - 3 cos (x) - 10 cot (x) + 6 = 0

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

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

y, cos (y) = x

\frac{sin ^{2} ( x )}{1 + cos ( x )} = 1

What is the general solution for 2cos^2(2x)-1=0 ?
The general solution for 2cos^2(2x)-1=0 is x= pi/8+pin,x=pi-pi/8+pin,x=(3pi)/8+pin,x=-(3pi)/8+pin