Помогите решить уравнение! (2sin x - √2)(2cosx + 1) = 0, где x принадлежит [-П; П]

(2sin x - √2)(2cosx + 1) = 0⇒ 1) 2sin x - √2=0⇒2sin x=√2⇒sinx=√2/2⇒ x=π/4+2πn; x=3π/4+2πn n=0⇒x=π/4∈[-π;π]; x=3π/4∈[-π;π] n=1⇒x=π/4+2π=9π/4∉[-π;π]; x=3π/4+2π=11/4∉[-π;π] n=-1⇒x=π/4-2π=-3π/4∈[-π;π]; x=3π/4-2π=-π/4∈[-π;π] n=-2⇒x=π/4-4π=-15π/4∉[-π;π]; x=3π/4-4π=-13/4∉[-π;π] 2) 2cosx + 1=0⇒2cos x=-1⇒cosx=-1/2⇒ x=2π/3+2πn; x=-2π/3+2πn n=0⇒x=2π/3∈[-π;π]; x=-2π/3∈[-π;π] n=1⇒x=2π/3+2π=8π/3∉[-π;π]; x=-2π/3+2π=8/3∉[-π;π] n=-1⇒x=2π/3-2π=-4π/3∉[-π;π]; x=-2π/3-2π=-8π/3∉[-π;π] Ответ: x=π/4; x=3π/4 x=-3π/4; x=-π/4 x=2π/3; x=-2π/3