What is the general solution for (4sin(2x)+4cos(2x))^2=16 ?

The general solution for (4sin(2x)+4cos(2x))^2=16 is x=pin+pi/2 ,x=pin+(3pi)/4 ,x=pin,x=pin+pi/4

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(4sin(2x)+4cos(2x))^2=16

(4 sin (2 x) + 4 cos (2 x))^{2} = 16

x = πn + \frac{π}{2}, x = πn + \frac{3 π}{4}, x = πn, x = πn + \frac{π}{4}

Solution steps

(4 sin (2 x) + 4 cos (2 x))^{2} = 16

Subtract

16

from both sides

(4 sin (2 x) + 4 cos (2 x))^{2} - 16 = 0

Factor

(4 sin (2 x) + 4 cos (2 x))^{2} - 16 : 16 (sin (2 x) + cos (2 x) + 1) (sin (2 x) + cos (2 x) - 1)

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

Solving each part separately

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

sin (2 x) + cos (2 x) + 1 = 0 : x = πn + \frac{π}{2}, x = πn + \frac{3 π}{4}

sin (2 x) + cos (2 x) - 1 = 0 : x = πn, x = πn + \frac{π}{4}

Combine all the solutions

x = πn + \frac{π}{2}, x = πn + \frac{3 π}{4}, x = πn, x = πn + \frac{π}{4}

2 sec (x) tan (x) + sec^{2} (x) = 0

sec^{2} (x) + 4 tan (x) = 6

sin^{2} (θ) = 4 - 2 cos^{2} (θ)

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

arcsin (x) = \frac{π}{7}

What is the general solution for (4sin(2x)+4cos(2x))^2=16 ?
The general solution for (4sin(2x)+4cos(2x))^2=16 is x=pin+pi/2 ,x=pin+(3pi)/4 ,x=pin,x=pin+pi/4