Recall the identity
#sin^2(x)+cos^2(x)=1#
Rewriting a bit, we get
#cos^2(x)=1-sin^2(x)#
So, we can rewrite our equation in terms of only #sin(x)#:
#sin(x)=1+(1-sin^2(x))#
#sin(x)=2-sin^2(x)#
Let's move everything to the left side and have the right side be #0:#
#sin^2(x)+sin(x)-2=0#
This looks like a quadratic equation, and we can treat it as one. The difference is, instead of having #x#, we have #sin(x).#
Let's factor this just as we would factor any other quadratic:
#(sin(x)+2)(sin(x)-1)=0#
Solving, we get
#sin(x)=-2#
#sin(x)=1#
Now, we need to solve each of the above equations for #x:#
#sin(x)=-2#
There is no solution to this. The range of #sin(x)# is #-1<=x<=1#, looking either at the graph or unit circle tells us this, as the function oscillates forever between #-1# and #1# . #sin(x)# can never be #>1# or #<-1#.
#sin(x)=1#
Let's consider where #sin(x)# is equal to #1#.
We know #sin(pi/2)=1#, so #x=pi/2# is certainly a valid answer. However, #x=(5pi)/2# is also a valid answer, as #(5pi)/2# means one and a half counterclockwise rotations around the unit circle, which would cause us to end up again at #pi/2#. The same applies for #(9pi)/2#, two and a half counterclockwise rotations around the unit circle, or #x=-(3pi)/2#, one and a half clockwise rotations around the unit circle.
Basically, adding or subtracting #2pi# (or any multiple of #2pi#) to #pi/2# gives us a value which causes #sin(x)# to equal #1#. This is due to the periodic nature of #sin(x)#. The period of #sin(x)# is #2pi#, so the value for #sin(x)# at a certain value of #x# repeats every #2pi# units .
So, we can denote our answer by
#x=pi/2+2pin# where n is any integer, thereby accounting for every possible answer.
