5tanx = 2sinsx solve for interval (0 2π)

Question

5tanx = 2sinsx solve for interval (0 2π)

algebraic means without calculator

Answer 1

5tanx = 2sinx

 

5(sinx/cosx) = 2sinx

 

5sinx = 2sinxcosx

 

5sinx - 2sinxcosx = 0

 

sinx(5-2cosx) = 0

 

sinx = 0 or cosx = 2.5

 

Since cosx always lies in the interval [-1,1], the equation cosx = 2.5 has no solution.  

 

So, sinx = 0.  x = 0, ±π, ±2π, ...

 

Solutions in the interval (o, 2π):  x = π

 

Answer 2

move it aside so it take form like this :

5tanx - 2sinx = 0

integrate it so it take like this 

5*ln|secx| + 2cosx for interval (0,2pi)

5* ln|sec0| + 2*cos0 - (5*ln|sex2pi| + 2cos2pi)

sec 0 =1 and ln 1 = 0 so 5*0 is 0

cos 0 = 1 so 2*1 = 1

0 and 2pi is the same in the trigonometry curve so the ans is

0 + 1-(0+1) = 0