secx +cosecx =c has four real roots between 0 and 2pi , then c^2>8?

Question

how????????????

Answer 1

first of all :
Sec X = 1 / Cos X
Cosec X = 1 / Sin X
Sec X + Cosec X = (Sin X + Cos X ) / ( Sin X . Cos X) .
For the equation you told : C should be less than √2 /2.
and so X should'nt be Pi/5.
for every single number of C <2√2 you 'll have 4 answers.
(one is in 1st quarter and one is in third quarter
OR
one in second quarter and one is in 4th quarter,).

Answer 2

secx +co secx =c
Squaring both side
sec^2 x +co sec^2 x + 2 (sec x) (co sec x)=c^2
Since sec x =1/cos x and cosec x = 1/sinx
1/cos^2 x + !/sin^2 x +2/sin x cos x = c^2
{sin^2x +cos^2x +2sin x cos x}/sin^x cos^2 x =c2
1+2sin x cos x}/sin^x cos^2 x =c2
Since 2sin x cos x =sin 2x, therefore
(1+ sin 2x)/ (sin^2 2x/4) =c^2
4(1+ sin 2x)/ (sin^2 2x) =c^2
Put sin 2x =t since sin 2x <1 0r 1/sin 2x >1 or 1/t >1
4(1+t)/t^2 =c2
4(1+t)/t^2 =c2
4(1/t^2 +1/t) =c2
4(>1 + >1) = c^2
4(>2) =c^2
Or c^2>8 hence proved