What is the general solution for sin(2x)-sqrt(2)cos(x)=sqrt(2)sin(x)-1 ?

The general solution for sin(2x)-sqrt(2)cos(x)=sqrt(2)sin(x)-1 is x=(3pi)/4+pin,x=2pin+pi/4

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sin(2x)-sqrt(2)cos(x)=sqrt(2)sin(x)-1

sin (2 x) - 2 cos (x) = 2 sin (x) - 1

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

Solution steps

sin (2 x) - 2 cos (x) = 2 sin (x) - 1

Subtract

2 sin (x) - 1

from both sides

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

cos^{2} (θ) - \frac{3}{2} cos (2 θ) = 0

sinh (x) = \frac{3}{4}, cosh (x)

tan (θ) = 1.4739716

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

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0 \leq x < 2 π

What is the general solution for sin(2x)-sqrt(2)cos(x)=sqrt(2)sin(x)-1 ?
The general solution for sin(2x)-sqrt(2)cos(x)=sqrt(2)sin(x)-1 is x=(3pi)/4+pin,x=2pin+pi/4