Solve for all values of x in the range 0° to 360°

7sin2x = cosx

Ive got as far as using a common trig identity of sin2A = 2sincosA

7(2sinθ cosθ) = cosθ

14sinθcosθ = cosθ

14sinθcosθ - cosθ = 0

cosθ(14sin-1)=0

either cosθ = 0

or 14sinθ -1 = 0

sinθ = 1/14 = 0.071

cosˆ-1(0) = 90

θ= 90,270

sinˆ-1(0.071) = 4.071

θ= 4.071 and 360-4.071 = 355.929

Could someone please confirm i have done this the correct way if not please advise.

Thanks in advance.