Question

2cos²x + 1 = 2√2cos (3π/2 - x)​

Answer 1

2cos²x + 1 = 2√2cos (3π/2 - x)​

2cos²x + 1 = -2√2sinx

2-2sin²x + 1 = -2√2sinx

2sin²x -3 -2√2sinx=0

sinx=√2±√(2+6)/2

а) sinx=√2+√(2+6)/2=(√2+2√2)/6=√2/2

x=(-1)ⁿarcsin(√2/2)+πn; n∈Z

x=(-1)ⁿπ/4+πn; n∈Z

б) sinx=√2-√(2+6)/2=(-√2)/2

x=(-1)ⁿarcsin(-√2/2)+πn; n∈Z

x=(-1)ⁿ⁺¹π/4+πn; n∈Z