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

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

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

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

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

Solution steps

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

Rewrite using trig identities

- cos^{2} (x) + 3 sin^{2} (x) = 0

Factor

- cos^{2} (x) + 3 sin^{2} (x) : (3 sin (x) + cos (x)) (3 sin (x) - cos (x))

(3 sin (x) + cos (x)) (3 sin (x) - cos (x)) = 0

Solving each part separately

3 sin (x) + cos (x) = 0 or 3 sin (x) - cos (x) = 0

3 sin (x) + cos (x) = 0 : x = \frac{5 π}{6} + πn

3 sin (x) - cos (x) = 0 : x = \frac{π}{6} + πn

Combine all the solutions

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

csc (x) cos^{2} (x) + sin^{2} (x) = csc (x)

0.87 = 5 cos (3 x)

cos (\frac{2 π x}{3}) = \frac{1}{2}

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

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

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