If x1 and x2 satisfy: 2sin(x)+sec(x)-2tan(x)-1=0, then sin(x1+x2)?

1 Answer
Oct 21, 2017

#2sin(x)+sec(x)-2tan(x)-1=0#

#=>2sin(x)cosx+sec(x)cosx-2tan(x)cosx-cosx=0#

#=>2sin(x)cosx+1-2sin-cosx=0#

#=>(1-cosx)-2sin(x)(1-cosx)=0#

#=>(1-cosx)(1-2sinx)=0#

So #cosx=1=>x=2npi->x_1#

And #sinx=1/2=>x=mpi+(-1)^mpi/6->x_2#
Where #n and m in ZZ#
So these are two values of x which satisfy the given equation.

Now

#sin(x_1+x_2)#

#=sinx_1cosx_2+cosx_1sinx_2#

#=sin(2npi)cos(mpi+(-1)^mpi/6)+cos(2npi)sin(mpi+(-1)^mpi/6)#

#=1/2#