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sinx-2-1-2-sin2x-2-cosx-2-0-x-0-2pi-

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Question Number 186685 by norboyev last updated on 08/Feb/23
(sinx)^2 −(1/2)sin2x−2(cosx)^2 ≥0 x∈[0;2π]
(sinx)2−12sin2x−2(cosx)2≥0x∈[0;2π]
Answered by Ar Brandon last updated on 08/Feb/23
sin^2 x−(1/2)sin2x−2cos^2 x≥0 ⇒sin^2 x−sinxcosx−2cos^2 x≥0 ⇒s^2 −s(√(1−s^2 ))−2(1−s^2 )≥0 ( ∣s∣ ≤1 ) ⇒3s^2 −2≥s(√(1−s^2 )) ⇒ 9s^4 −12s^2 +4≥s^2 (1−s^2 ) ⇒10s^4 −13s^2 +4≥0 ⇒ (2s^2 −1)(5s^2 −4)≥0 ⇒((√2)s−1)((√2)s+1)((√5)s−2)((√5)s+2)≥0 ⇒s∈(−1; −(2/( (√5)))]∪[−(1/( (√2))); (1/( (√2)))]∪[(2/( (√5))); 1) ⇒x∈[−(π/2); −arcsin((2/( (√5))))]∪[−arcsin((2/( (√5))));arcsin((2/( (√5))))] ∪[arcsin((2/( (√5)))); (π/2)]
sin2x−12sin2x−2cos2x⩾0⇒sin2x−sinxcosx−2cos2x⩾0⇒s2−s1−s2−2(1−s2)⩾0(∣s∣⩽1)⇒3s2−2⩾s1−s2⇒9s4−12s2+4⩾s2(1−s2)⇒10s4−13s2+4⩾0⇒(2s2−1)(5s2−4)⩾0⇒(2s−1)(2s+1)(5s−2)(5s+2)⩾0⇒s∈(−1;−25]∪[−12;12]∪[25;1)⇒x∈[−π2;−arcsin(25)]∪[−arcsin(25);arcsin(25)]∪[arcsin(25);π2]

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