2sin2x+cosx=0

Question

can u please try to find q1 2 3 nd 4

Answer 1

sin2x=2cosxsinx, so substitute, and you get 4sinxcosx+cosx=0. 

 

Factor out the cosx, and you get cosx(4sinx+1)=0, so solve for

 

cosx=0 and

sinx=-1/4

Answer 2

2sin(2x) + cosx = 0

 

2(2sinxcosx) + cosx = 0

 

cosx(4sinx + 1) = 0

 

cosx = 0 or sinx = -1/4

 

Here are the solutions in the interval 0 ≤ x ≤ 2π:

 

           cosx = 0 when x = π/2, 3π/2

 

           If sinx = -1/4, then x lies in Q3 or Q4

 

                     sin^-1(1/4) = .253

 

                     So, x = π + .253 or 2π - .253

 

                               = 3.395, 6.030