Question

Buna! Stie cineva?
4 sin²x-2cosx=4-√2(2cos x +1)

Answer 1

4sin^2x= 4-4cos^2 x. (1=sin^2x+cos^2x)
4-4cos^2x-2cosx=4-2√2cosx-√2
cosx=t
-4t^2+(2√2-2)t+√2=0
∆=8-8√2+4+16√2=8+8√2+4=(2√2+2)^2
t1=(-2√2+2+2√2+2)/-8=-1/2
t2=(-2√2+2-2√2-2)/-8=-√2/2
cosx=-1/2 x=2π/3+ 2kπ , x=4π/3+2kπ
cosx=-√2/2. x= 3π/4+2kπ. , x=5/4π+2kπ