Question

Помогите,пожалуйста!!! найти область значений функции y=1/(sin^6(x)+cos^6(x))

Answer 1

$1=1^3=(sin^2x+cos^2x)^3=\\\\=sin^6x+cos^6x+3sin^2x\cdot cos^2x(sin^2x+cos^2x)=\\\\=sin^6x+cos^6x+3(\frac{1}{2}sin2x)^2=sin^6x+cos^6x+\frac{3}{4}sin^22x\; \to \\\\sin^6x+cos^6x=1-\frac{3}{4}sin^22x\\\\\\f(x)= \frac{1}{sin^6x+cos^6x} = \frac{1}{1-\frac{3}{4}sin^22x} \\\\0 \leq sin^22x \leq 1\\\\ -\frac{3}{4}\leq -\frac{3}{4}sin^22x \leq 0\; ,\; \; \frac{1}{4} \leq 1-\frac{3}{4}sin^22x \leq 1\\\\1 \leq \frac{1}{1-\frac{3}{4}sin^22x} \leq 4\\\\f(x)\in [\, 1,4\, ]$