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

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

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

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

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

Solution steps

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

Subtract

4

from both sides

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

Factor

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

4 (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}

sin (x) = \frac{7}{3}

0 = sin (4 x)

x, sin (x) = sin (2 x)

- 7 tan (A) - 14 = tan (A) - 6

5 csc (θ) - 8 = 0

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