Question

а) cos 2x -√3 sin (П/2 -х) +1=0
б) укажите корни этого уравнения, принадлежащие отрезку
[ -4П;  5П/2]

Answer 1

$cos (2x) - sqrt{3}sinBig(dfrac{pi}2-xBig)+1=0\\cos (2x) - sqrt{3}cos x+1=0\\2cos^2x -1 - sqrt{3}cos x+1=0\\2cos^2x - sqrt{3}cos x=0\\ cos x(2cos x-sqrt{3})=0\\1)~~cos x=0;~~~x_1=dfrac{pi}2+pi n,~~n in Z\\2)~~2cos x-sqrt{3}=0; ~~cos x = dfrac{sqrt3}2\\~~~~~x_2=dfrac{pi}6+2pi k, ~~x_3=-dfrac{pi}6+2pi m,~~k,m in Z$

Выбор корней на интервале  [-4π; 5π/2] в приложении

[-4π; 5π/2]    ⇔    [-4π; 2,5π]

$x_1=dfrac{pi}2+pi n,~~n in {-4; -3; -2; -1; 0; 1; 2}\\ boxed{boldsymbol{-3,5pi;~-2,5pi;~-1,5pi;~-0,5pi;~0,5pi;~1,5pi;~2,5pi}} \\\x_2=dfrac{pi}6+2pi k, ~~k in {-2; -1; 0; 1}\\ boxed{boldsymbol{Big(-3dfrac56Big)pi;~Big(-1dfrac56Big)pi;~Big(dfrac16Big)pi;~Big(2dfrac16Big)pi}} \\\x_3=-dfrac{pi}6+2pi m, ~~m in {-1; 0; 1}\\ boxed{boldsymbol{Big(-2dfrac16Big)pi;~Big(-dfrac16Big)pi;~Big(1dfrac56Big)pi}}$