Question

 

4cos^2x+9sinx=5

^2-квадрат

 

 

 

 

 

Answer 1

$4cos^2 x +9 sin x = 5\ 4-4sin^2 x +9 sin x = 5\ 4sin^2 x - 9 sin x +1 = 0\ D = sqrt{(81 - 16)} = sqrt{65}\ sin x_{1,2} = frac{9 pm sqrt{65}}{8}\ frac{9 +sqrt{65}}{8}>1\ sin x = frac{9 -sqrt{65}}{8}\ x = arcsin{(frac{9 -sqrt{65}}{8})}$