is this right ?
i wonder if this is right :)
4 sin^2(x) + 8 cos(x) = 7
1 = cosē(x)+sinēx
sinē(x) = 1-cosē(x)
replace sinē(x) with 1-cosē(x)
-4cosē(x)+8cos(x)+4 = 7
replace [cos(x)=t] and extender with-1
4tē-8t-4 = -7 <--> t1= 1/2 t2=3/2
then you get the first x but the other ?
cos(x2) = 3/2
Is not possible for real values of x2, so this does not give any solutions.
but the solution is right ?
Looks like it.
Originally Posted by paulaa
Solutions are all x s.t. cos(x) = 1/2.
Careful though, since there is an infinite family of such values x.