Question

14. 4 -/1 points McKTrig8 6.2.037. Find all degree solutions in the interval 0° SO < 360°. If rounding is necessary, 10 cos + 13 tan 6 = sec Need Help? Read It Talk to a Tutor

Answer 1

110 COS 0+ 13 jane = seco where o is ooco< 360° we know quotient identity: hana sine Reciprocal identity seco = 1 Pythogorean identity sinzo Hos?o=1 Cose Coso os?0 = 1 - sino Now question is Jocosat 13 tano = seco Llocoso + 13 Sino al coso caso multiply both side by coso L10 cos²o + 13 sino = L 10(1-sino) + 13 sino = 1 Lio - losin²o + 13 sino-1=0 - 10 sen?o +13 sino t0-150 -10 sin?o +13 Sino +9=0 Taking - as common both side 10 din? - 13 sino -9.20 .110x9=90 lo sino - 18 fino +5 sino -9:0 | 1845390 asino (5 sino -9) +15 Sino-920 / 18-5=13 - (5 sino-2) (asinotl=0 15.Sino920- _5&ne age sino = 9 = 1.8 since sinio vary from 1 to L Hence sino = 1.8 is not define Therefore sino=1.8 does not exit. 2 Sino +1 -0 2 Sino-1 Sinoz-

sino=-1 sino = sin(-30°) since sino is negative in quardant [II, IV Hence sin(180° +o) = sin(o) - sin (360°-o) = sin(-o) sino = sin(-30°) sin (o) = sin(180° +30°) . o = 210° - Again sino = sin(-30°) sin (o) = sin (360° -30) sin(0) = sin(3300) o = 3300 Hence o = 210°, 330° in jutegral O LO <360°