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

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

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

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

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

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

cos (2 x) = 1 : x = πn

Combine all the solutions

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn, x = πn

4 cos (x) + 2 = 3

sin (4 θ) = 4 sin (2 θ) cos (2 θ)

cos (x) = 6

6 sin (x) - 1 = 4 sin (x)

sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

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