Question

2sin (7π/2+x)* sin x =√3 cos x, [-7π;-6π]

Answer 1

2sin(7π/2 + x)*sinx = √3*cosx
2sin(π/2 + x)sinx = √3cosx
2cosxsinx - √3cosx = 0
cosx(2sinx - √3) = 0
1)  cosx = 0
x₁ = π/2 + πk, k∈Z
2)  2sinx - √3 = 0
sinx = √3/2
x = (-1)^(n)*arcsin(√3/2) + 2πn, n∈Z
x₂ = (-1)^(n)*(π/3) + 2πn, n∈Z