Solve `sin x("cos"x/4-2 sin x)+(1+"sin"x/4-2 cos x)cos x =0`.

Answer

`x=(8k-6) pi, k in Z`

Explanation

`sin x ("cos" x/4-2 sin x)+(1+ "sin" x/4-2 cos x) cos x -0`
or `(sin x "cos"x/4+ "sin" x/4 cos x) -2 (sin^(2) x+ cos^(2) x)+ cos x=0`
or `"sin" (5 x)/4 + cos x =2` ...(i)
Now, Eq. (i) will hold if
`"sin" (5x)/4=1` and `cos x =1`
`rArr (5x)/4=2npi + pi/2, n in Z and x=2m pi, m in Z`
`rArr x=((8n+2)pi)/5, n in Z` ...(ii)
and `x=2mpi, m in Z` ...(iii)
Comparing the values of x, we get
`((8n+2)pi)/5=2 mpi`
or `8n +2 =10 m`
or `n=(5m-1)/4`

Therefore, the general solution of the given equation can be obtained by substituting either `m=4k-3` in Eq. (iii) or `n=5p-4` in Eq. (ii).
Thereore, the general solution of Eq. (i) is
`x=(8k-6) pi, k in Z`.