Question

tan^(2)2x+cot^(2)2x+2tan2x+2cot2x=6

Answer 1

Given : Tan²(2x) + Cot²(2x)  + 2tan(2x)  + 2Cot(2x) = 6

To find : number of solutions of equation in  x ∈ (0,pi)

Solution:

Tan²(2x) + Cot²(2x)  + 2tan(2x)  + 2Cot(2x) = 6

=> Tan²(2x) + Cot²(2x) + 2 + 2tan(2x)  + 2Cot(2x) = 6 + 2

=> Tan²(2x) + Cot²(2x) + 2 Tan²(2x)Cot²(2x)  + 2tan(2x)  + 2Cot(2x) = 8

=> ( Tan(2x) + Cot(2x)) ²  + 2 ( Tan(2x) + Cot(2x) ) = 8

=> ( Tan(2x) + Cot(2x)) ²  + 2 ( Tan(2x) + Cot(2x) ) -  8   = 0

Let say Tan(2x) + Cot(2x) = y

=> y²  + 2y - 8 = 0

=> y² + 4y - 2y - 8 = 0

=> y(y + 4) - 2(y + 4) = 0

=> (y - 2)(y + 4) =  0

=> y = 2 or y = -4

Tan(2x) + Cot(2x)  = 2

=> Tan(2x) +  1/Tan(2x)  = 2

=> Tan²(2x)  - 2Tan(2x) + 1  = 0  

=> ( Tan(2x) - 1)² = 0

=> Tan(2x) - 1) = 0

=> Tan(2x) = 1  

=> 2x  =    π/4  , 5π/4

=> x = π/8  , 5π/8

Tan(2x) + Cot(2x)  = -4

=> Tan(2x) +  1/Tan(2x)  = 4

=> Tan²(2x)  + 4Tan(2x) + 1  = 0  

Tan(2x)  =( - 4 ± √12)/2

= - 2 ± √3

Tan(2x)    = - 2 + √3    ,  - 2 - √3  

2x =  two values in ( π/2 , π )  & two values in ( 3π/2 , 2π )

4 values of  x  in ( 0 , π)

Total 6 possible solution  in ( 0 , π ) for x

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