Question

slove the following questions​

Answer 1

by takin thita as x

sin^2 x+2 cos x +1/4 =0

1- cos^2 x +2cos x + 1/4 =0

cos ^2 x -2cos x _5/4=0

cos x =( 2±✓4-4*1*-5/4)/2

cos x = (2±✓4+5)/2

cos x = (2±3)/2

either cos x = 5/2 or cos x = -1/2

since cos value lie between -1 to +1

rjerefore 5/2 can be neglected

hence we get

cos x =-1/2

or x=2π/3 or x= 4π/3

AND BRO SORRY FOR PREVIOUS ANSWER