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(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