√(3+2sin^2 x) =√(6cos 0,5x)
3+2sin^2 x=6cos^2 0,5x; 2cos^2 x/2=1+cosx
3+2*(1-cos^2 x=3*(1+cosx)
3+2-2cos^2 x-3-3cosx=0
-2cos^2 x -3cosx +2=0
2cos^2 x+3cosx-2=0
y=cosx; 2y^2 +3y -2=0
D=9-4*2*(-2)=9+16=25=5^2; y1=(-3-5)/4=-2; y2=(-3+5)/4=1/2
cosx=-2 ili cosx=1/2
корней нет x=+- π/3 +2πn; n-celoe
Проверка: х=π/3; √(3+2sin^2 π/3) =√6 *cos π/3 неверно
√(3+2*(√3 /2)^2 )=√6 *(1/2)
√(3+2(3/4)=√(3+3/2)=√9/2=3√2/ 2
x=-π/3; √(3+2*(-√3/2)^2) =√(3+2* 3/4)=√(3+3/2)=3√2 /2
√6 *cos(-π/3)=√6 cosπ/3=√6 *(1/2=√6 /2
Ответ решений нет