Question

2sin^2 ×X-cosX-1=0 [3π;4π]

Answer 1

2(1-сos²x)-cosx-1=0
2-2cos²x-cosx-1=0
cosx=a
2a²+a-1=0
D=1+8=9
a1=(---3)/4=-1⇒cosx=-1⇒x=π+2πn,n∈z
3π≤π+2πn≤4π
3≤1+2n≤4
2≤2n≤3
1≤n≤1,5
n=1⇒x=π+2π=3π
a2=(-1+3)/4=1/2⇒cosx=1/2⇒x=+-π/3+2πn
3π≤-π/3+2πn≤4π
9≤-1+6n≤12
10≤6n≤13
10/6≤n≤13/6
n=2⇒x=-π/3+4π=11π/3
3π≤π/3+2πn≤4π
9≤1+6n≤12
8≤6n≤11
8/6≤n≤11/6
нет решения