Find the largest value of the function y = 10cos ^ 2-6sincos + 2sin ^ 2.

1. Let’s transform the function:

y = 10cos ^ 2x – 6sinx * cosx + 2sin ^ 2x;
y = (3cosx – sinx) ^ 2 + 1;
y = (√10 (3 / √10 * cosx – 1 / √10 * sinx)) ^ 2 + 1;
y = 10 (3 / √10 * cosx – 1 / √10 * sinx) ^ 2 + 1.
2. Let:

φ = arccos (3 / √10).

Then:

cosφ = 3 / √10;
sinφ = 1 / √10.
We get the function:

y = 10 (cosφ * cosx – sinφ * sinx) ^ 2 + 1;
y = 10cos ^ 2 (x + φ) + 1.
3. For the cosine function, the following inequality is true:

cos (x + φ) ≤ 1, hence:
cos ^ 2 (x + φ) ≤ 1;
10cos ^ 2 (x + φ) ≤ 10;
10cos ^ 2 (x + φ) + 1 ≤ 11;
y ≤ 11.
Highest function value: 11.

Answer: 11.