Ответы на вопрос:
2sin²x+2cos²2x-1=0 2(1-cos2x)/2+2cos²2x-1=0 2cos²2x-cos2x=0 cos2x(2cos2x-1)=0 cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2 5π/2≤π/4+πn/2≤3π 10≤1+2n≤12 9≤2n≤11 4,5≤n≤5,5 n=5 x=π/4+5π/2=11π/4 cos2x=1/2⇒2x=+-π/3+2πn⇒x=+-π/6+πn 5π/2≤-π/6+πn≤3π 15≤-1+6n≤18 16≤6n≤19 2 2/3≤n≤3 1/6 n=3 x=-π/6+3π=17π/6 5π/2≤π/6+πn≤3π 15≤-1+6n≤18 14≤6n≤17 2 1/3≤n≤2 5/6 нет решения ответ x=11π/4 u x=17π/6
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