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

The general solution for 4cos^2(2x)-2=0 is x= pi/8+pin,x=pi-pi/8+pin,x=(3pi)/8+pin,x=-(3pi)/8+pin

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

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

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

tan (θ) = \frac{9}{12}

tan (a) = \frac{20}{21}

sin (\frac{2 x}{3}) = - 1

3 tan (θ) sin (θ) - 2 tan (θ) = 0

- 8 csc^{2} (x) + 2 cot (x) = - 23

What is the general solution for 4cos^2(2x)-2=0 ?
The general solution for 4cos^2(2x)-2=0 is x= pi/8+pin,x=pi-pi/8+pin,x=(3pi)/8+pin,x=-(3pi)/8+pin