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

The general solution for 2cos^2(2x)+sin(2x)=1 is x=(7pi)/(12)+pin,x=(11pi)/(12)+pin,x= pi/4+pin

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

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

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{π}{4} + πn

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (2 x) = - \frac{1}{2} : x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn

sin (2 x) = 1 : x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{7 π}{12} + πn, x = \frac{11 π}{12} + πn, x = \frac{π}{4} + πn

sin (4 θ) + sin (2 θ) = 0

sec (x) = \frac{2}{3}

2 sin (x) + 1 = 3 csc (x)

14 sin (20 x) - 3 = 2

3 tan (3 x) - 1 = 0

What is the general solution for 2cos^2(2x)+sin(2x)=1 ?
The general solution for 2cos^2(2x)+sin(2x)=1 is x=(7pi)/(12)+pin,x=(11pi)/(12)+pin,x= pi/4+pin