What is the general solution for cos(2x)=1-sin(2x),0<= x<= 2pi ?

The general solution for cos(2x)=1-sin(2x),0<= x<= 2pi is x=0,x= pi/4 ,x=pi,x=(5pi)/4 ,x=2pi

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cos(2x)=1-sin(2x),0<= x<= 2pi

cos (2 x) = 1 - sin (2 x), 0 \leq x \leq 2 π

x = 0, x = \frac{π}{4}, x = π, x = \frac{5 π}{4}, x = 2 π

Solution steps

cos (2 x) = 1 - sin (2 x), 0 \leq x \leq 2 π

Subtract

1 - sin (2 x)

from both sides

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

Rewrite using trig identities

- 1 + 2 sin (2 x + \frac{π}{4}) = 0

Move

1

to the right side

2 sin (2 x + \frac{π}{4}) = 1

Divide both sides by

2

sin (2 x + \frac{π}{4}) = \frac{2}{2}

General solutions for

sin (2 x + \frac{π}{4}) = \frac{2}{2}

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

Solve

2 x + \frac{π}{4} = \frac{π}{4} + 2 πn : x = πn

Solve

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

x = πn, x = πn + \frac{π}{4}

Solutions for the range

0 \leq x \leq 2 π

x = 0, x = \frac{π}{4}, x = π, x = \frac{5 π}{4}, x = 2 π

0 = 1 + 2 sin (θ)

csc (x - \frac{π}{3}) = 2

sin (θ) = \frac{11}{6}

x, 5 cos (5 x) = 0

sin (3 x) - sin (5 x) = 0

What is the general solution for cos(2x)=1-sin(2x),0<= x<= 2pi ?
The general solution for cos(2x)=1-sin(2x),0<= x<= 2pi is x=0,x= pi/4 ,x=pi,x=(5pi)/4 ,x=2pi