Cos4x -sin2x =0 ; x ∈ [ 0 ;π] .
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1 -2sin²2x - sin2x =0 ⇔2sin²2x +sin2x -1 =0 ⇒ [ sin2x = -1 ; sin2x =1/2.
[ 2x = -π/2 +2πn ; 2x = π/6 + 2πn ; 2x = 5π/6 + 2πn , n∈ Z.⇔
[ x = -π/4 +πn ; x = π/12 + πn ; x = 5π/12 + πn , n∈ Z.
ответ: { π/12 ; 5π/12 ; 3π/4} .
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a) 0≤ -π/4 + πn ≤ π ⇔ π/4 ≤ πn ≤ 5 π/4 ⇔1/4 ≤ n ≤ 5 /4 ⇒ n = 1.
решение: -π/4 + π =3π/4 .
b) 0≤ π/12 + πn ≤ π ⇔ - π/12 ≤ πn ≤ π - π/12⇔-1/12 ≤ n ≤ 11/12 ⇒ n =0.
решение: π/12 + πn = π/12.
c) 0 ≤5π/12 + πn ≤ π ⇔ - 5π/12 ≤ πn ≤ π- 5π/12⇔ -5/12 ≤ n ≤ 71/12⇒ n =0
решение: 5π/12 + πn= 5π/12 .