Find solution of eqation 2cot^2x-5cosec x =1

[vkash]

2cot²(x)-5cosec(x)=1 for exactly 6 values of x belongs to [0,nπ], then find the minimum value of n.
my answer is 5.
How i did it.changing cot to cosec and then solving equation it will give something like cosec(x)=3. for x belongs to 0 to 5π it has 6 solutions..
this is question of a class test & answer in test's answer sheet is 6.
WHO IS correct. me or answer sheet??

 

[rock.freak667]

In 2π, you'd get 2 solutions A and 180-A. So for the next 2π, you'd have 4 in total and then in 2π again, you'd get 6.

So your range would be 6π.

 

[vkash]

In 2π, you'd get 2 solutions A and 180-A.[/color] So for the next 2π, you'd have 4 in total and then in 2π again, you'd get 6.

So your range would be 6π.

If i have got A and 180-A then i have got solutions in [0,π] not [0,2π] and in next 2π(or in [0,3π]) i will have 4 solution why i should go for 4π. similarly for next 5π.

 

[SammyS]

You are correct for the problem that was given.

There are two roots in [0, π] ,

two roots in [2π, 3π] ,

and two roots in [4π, 5π] .

 

[vkash]

You are correct for the problem that was given.

There are two roots in [0, π] ,

two roots in [2π, 3π] ,

and two roots in [4π, 5π] .

thanks sammy..