# The no. of solutions of equation tanx+secx=2cosx lying in the interval [0,2pi] is

Vikas TU
14149 Points
6 years ago
Dear Student,
tanx+sec x=2cosx => sin x/cos x + 1/cosx=2 cosx =>(sinx+1)/cosx = 2cosx => 1+sinx=2 cos^2 x
=>1+sin x=2(1-sin^2 x) =>1+sinx -2(1+sinx)(1-sinx)=0 => (1+sinx)(1-2(1-sinx))=0 =>(1+sinx)(2sinx-1)=0
so sin x=1, sin x=1/2
so in between [0,2pi], the solutions are x=3pi/2, x=pi/6, x=5pi/6.
therefore there are 3 solutions.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Pk
13 Points
5 years ago
But secx tan are not define in (2n+1)pi/2 so the ans will be 0 We can not multiply of cosx to other. Side in frist we follow the condition
Satya prakash Rawte
24 Points
5 years ago
Everything you have solved is correct, but the conclusion of 3 solutions is wrong.
As you are saying there are 3 solution, that means they should satisfy the original equation.
But x = 3pi/2 does not satisfy the equation.
therefore ans is 2 solution
CalebS
14 Points
3 years ago
tanx+sec x=2cosx
sin x/cos x + 1/cosx=2 cosx
(sinx+1)/cosx = 2cosx
1+sinx=2 cos^2 x
1+sin x=2(1-sin^2 x)
1+sinx -2{(1+sinx)(1-sinx)}=0
(1+sinx){1-2(1-sinx)}=0
(1+sinx)(2sinx-1)=0
So, sin x=1 & sin x=1/2
so in between [0,2π], the solutions are x=3π/2, x= π /6, x=5π /6
3 years ago
Dear student,

tanx+sec x=2cosx
sin x/cos x + 1/cosx=2 cosx
(sinx+1)/cosx = 2cosx
1+sinx=2 cos2x
1+sin x=2(1-sin2x)
1+sinx -2{(1+sinx)(1-sinx)}=0
(1+sinx){1-2(1-sinx)}=0
(1+sinx)(2sinx-1)=0
So, sin x=1 & sin x=1/2
so in between [0,2π], the solutions are x=3π/2, x= π /6, x=5π /6
But, at x = 3π/2, tanx and secx are not defined
Hence, the equation has oonly 2 solutions x= π /6 and x=5π /6

Hope it helps.
Thanks and regards,
Kushagra