Solve : `5cos2theta+2cos^2(theta/2)+1=0,-pi lt theta lt pi`

Solve 5 cos 2 theta+2 "cos"^(2) theta/2 +1=0, -pi lt theta lt pi .

Solve 5 cos 2 theta+2 "cos"^(2) theta/2 +1=0, -pi lt theta lt pi .

Solve: 5cos2 theta+2cos^(2)((theta)/(2))+1=0,-pi

Solve the following equation (i) 5cos2theta+2cos^(2)"(theta)/(2)+1=0 , -(pi)/(2) lt theta lt (pi)/(2) (ii) sin7theta+sin4theta+sintheta=0 , 0 le theta le pi (iii) tantheta+sectheta=sqrt(3) , 0 le theta le 2pi

Solve the following equation (i) 5cos2theta+2cos^(2)"(theta)/(2)+1=0 , -(pi)/(2) lt theta lt (pi)/(2) (ii) sin7theta+sin4theta+sintheta=0 , 0 le theta le pi (iii) tantheta+sectheta=sqrt(3) , 0 le theta le 2pi

If 5 cos 2 theta + 2 cos ^(2) ""(theta)/(2) + 1 = 0 , - pi lt theta lt pi then number of solutions are

sin 2theta = cos 3theta (0 lt theta lt pi)