2sin^2 ×X-cosX-1=0 [3π;4π]

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