If `5 cos 2theta + 2 cos^2 (theta/2) + 1 = 0`,`-pi < theta < pi` , then `theta` can be

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

If 5 cos 2 theta +2 cos^2 (theta/2)+1=0 , when (0 ltthetalt pi) , then the values of theta are :

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

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

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