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 cos x,

cosx2sinx−1=0 

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

cos x=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+2n and x=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 ( n an arbitrary integer).

√ 2sin x−1=0 

if we rearrange the equation, we obtain the basic equation as

sin x =12 , which has the solutions x=4 and x=34 in the unit circle and hence the general solution

x=4+2n and x=43+2n

where n can arbitrary integer.

All in all, the original equation has the solutions

x=4+2nx=2+nx=43+2n ( n an arbitrary integer).