Найдите значение выражения :5cos 2x+6sin²x , если ctgx = 1/3

$5 \cos 2x + 6\sin^{2}x = 5(\cos^{2}x - \sin^{2}x) + 6\sin^{2}x =\\= 5\cos^{2}x - 5\sin^{2}x + 6\sin^{2}x = 5\cos^{2}x + \sin^{2}x =\\= 4\cos^{2}x + \cos^{2}x + \sin^{2}x = 4\cos^{2}x + 1$

$1 + \text{ctg}^{2} \, x = \dfrac{1}{\sin^{2}x} = \dfrac{1}{1 - \cos^{2}x}$

$1 - \cos^{2}x = \dfrac{1}{1 + \text{ctg}^{2} \, x}$

$\cos^{2} x = 1 - \dfrac{1}{1 + \text{ctg}^{2} \, x} = 1 - \dfrac{1}{1 + \left(\dfrac{1}{3} \right)^{2}} = 1 - \dfrac{1}{1 + \dfrac{1}{9} } = 0,1$

$5 \cos 2x + 6\sin^{2}x= 4 \cdot 0,1 + 1 = 1,4$

Ответ: $1,4$