Question

1)sinx+cosx=1

2)sinx+cosx=1+sin2x

Answer 1

Are we solving ? I will assume domain 0 ≤ x≤ 2π

first one:
square both sides
sin^2 x + 2sinxcosx + cos^2 x = 1
1 + sin 2x = 1
sin2x = 0
2x = 0 , π, 2π , 3π, 4π
x = 0 , π/2, π , 3π/2 2π
but since we squared , we have to verify all answers
for x=0
LS = sin) + cos) = 0+1 = 1 = RS
for x=π/2
sinπ/2 + cosπ/2 = 1/√2 + 1/√2 = 2/√2 ≠ RS
for x= π
sinπ + cosπ = 0-1 ≠ RS
for x= 3π/2
LS = .......≠ RS
for x=2π
LS = sin2π + cos2π = 0 + 1 = 1 = RS

so x = 0 or x = 2π

Answer 2

also square both sides
sin^2 x + 2sinxcosx + cos^2 x = 1 + 2sin2x + sin^2 (2x)
2sinxcosx = 2sin2x + sin^2 (2x)
sin 2x = 2 sin 2x + sin^2(2x)
sin^2(2x) + sin2x = 0
sin2x(sin2x + 1) = 0
sin2x = 0 or sin2x = -1
2x = 0,π,2π,3π,4π or 2x = 3π/2, 7π/2

x = 0,π/2,π,3π/2,2π or x = 3π/2,7π/4

The above are <b>potential</b> solutions, I will leave it up to you to verify which are actual solutions.
Remember, we squared , so all solutions must be verified.