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

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

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

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

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

Solution steps

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

Subtract

cos (2 x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{2}{2}, sin (x) = 0

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

sin (x) = 0 : x = 2 πn, x = π + 2 πn

Combine all the solutions

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

cos (x) = 0.89

cos (x) = 0.88

cot (\frac{θ}{2}) = - 1

sin (2 θ) cos (θ) - sin (θ) = 0

- 12 sin (x) + 6 sin (2 x) = 0

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