Solve the following equation for 0^{0} ≤ x ≤ 360^{0} 2cosx = sin^{2} x + 2

2 cos x = 1 - cos^{2}x + 2cos^{2}x + 2cosx - 3 = 0p = cos xp^{2} + 2p -3 = 0p^{2} + 3p -p -3 =0p(p+3) - 1(p+3) =0 p = 1 or -3cos x = 1x=0º, 360º

5.9k questions

8.3k answers

6 comments

590 users