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

The general solution for (sin(2x)+cos(2x))^2=2 is x=pin+(5pi)/8 ,x=pin+pi/8

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

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

x = πn + \frac{5 π}{8}, x = πn + \frac{π}{8}

Solution steps

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

Subtract

2

from both sides

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

Factor

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

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

Solving each part separately

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

sin (2 x) + cos (2 x) + 2 = 0 : x = πn + \frac{5 π}{8}

sin (2 x) + cos (2 x) - 2 = 0 : x = πn + \frac{π}{8}

Combine all the solutions

x = πn + \frac{5 π}{8}, x = πn + \frac{π}{8}

1.5 = 6 sin (x)

\frac{2 sin ( x ) - 1}{3 \cdot cos ( x )} = 0

tan (θ) = \frac{34400}{254}

2 sin (x) = - 1, - π \leq x \leq π

cot (- \frac{π}{5}) - sec (x) = 1.5

What is the general solution for (sin(2x)+cos(2x))^2=2 ?
The general solution for (sin(2x)+cos(2x))^2=2 is x=pin+(5pi)/8 ,x=pin+pi/8