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/2+pin,x= pi/6+pin,x=(5pi)/6+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}{2} + πn, x = \frac{π}{6} + πn, x = \frac{5 π}{6} + 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) = - 1, cos (2 x) = \frac{1}{2}

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

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

Combine all the solutions

x = \frac{π}{2} + πn, x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

2 \cdot cos (x) - 1 = 0

cos (x) + 3 sin (x) = 2

2 sin (θ) = - 2, 0 \leq θ < 2 π

y, x = arcsin (y^{2})

cos^{2} (2 x) + cos (2 x) - \frac{3}{4} = 0

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/2+pin,x= pi/6+pin,x=(5pi)/6+pin