Thread: is this right ?

i wonder if this is right

4 sin^2(x) + 8 cos(x) = 7

1 = cos²(x)+sin²x

sin²(x) = 1-cos²(x)

replace sin²(x) with 1-cos²(x)

-4cos²(x)+8cos(x)+4 = 7

replace [cos(x)=t] and extender with-1

4t²-8t-4 = -7 <--> t1= 1/2 t2=3/2

cos(x1)=1/2
cos(x2)=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 ?

Originally Posted by paulaa

but the solution is right ?

Looks like it.

Solutions are all x s.t. cos(x) = 1/2.

Careful though, since there is an infinite family of such values x.