Solution 4.4:6b

From Förberedande kurs i matematik 1

After moving the terms over to the left-hand side, so that

2sinxcosx−cosx=0

we see that we can take out a common factor cosx,

cosx(2sinx−1)=0

and that the equation is only satisfied if at least one of the factors cosx or 2sinx−1  is zero. Thus, there are two cases:

cosx=0:

This basic equation has solutions x=2 and x=32 in the unit circle, and from this we see that the general solution is

x=2+2nandx=23+2n

where n is an arbitrary integer. Because the angles 2 and 32 differ by , the solutions can be summarized as

x=2+n

where n is an arbitrary integer.

2sinx−1=0: 

If we rearrange the equation, we obtain the basic equation as sinx=12 , which has the solutions x=4 and x=34 in the unit circle and hence the general solution

x=4+2nandx=43+2n

where n is an arbitrary integer.

All in all, the original equation has the solutions

xxx=4+2n=2+n=43+2n

where n is an arbitrary integer.