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

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

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

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

x = πn, x = πn + \frac{π}{4}

Solution steps

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

4 tan (x) sin^{2} (x) + 3 = 4 sin^{2} (x) + 3 tan (x)

tan (a) = 0.384

cos (b) = \frac{38}{70}

cos (x) = 1 - sin^{2} (\frac{x}{2})

3 sin (x) - 4 sin^{3} (x) = 1 - 2 sin^{2} (x)

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