Question

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1 неравенство
задание на фото 

Answer 1

$sin^6x+cos^6x \geq \frac{5}{8}$

$(sin^2x)^3+(cos^2x )^3\geq \frac{5}{8}$

$(sin^2x+cos^2x )(sin^4x-sin^2xcos^2x+cos^4x)\geq \frac{5}{8}$

$(sin^2x)^2+(cos^2x)^2-sin^2xcos^2x\geq \frac{5}{8}$

$(sin^2x+cos^2x)^2-2sin^2xcos^2x-sin^2xcos^2x\geq \frac{5}{8}$

$(sin^2x+cos^2x)^2-3sin^2xcos^2x\geq \frac{5}{8}$

$1-3sin^2xcos^2x\geq \frac{5}{8}$

$-3sin^2xcos^2x\geq \frac{5}{8} -1$

$-3sin^2xcos^2x\geq - \frac{3}{8}$

$3sin^2xcos^2x \leq \frac{3}{8}$

$\frac{3}{4}* sin^22x \leq \frac{3}{8}$

$sin^22x \leq \frac{3}{8} * \frac{4}{3}$

$sin^22x \leq \frac{1}{2}$

$\frac{1-cos4x}{2} \leq \frac{1}{2}$

$1-cos4x} \leq 1$

$-cos4x} \leq 0$

$cos4x \geq 0$

$- \frac{ \pi }{2} +2 \pi n \leq 4x \leq \frac{ \pi }{2} +2 \pi n,$  $n$ ∈ $Z$

$- \frac{ \pi }{8} + \frac{ \pi n}{2} \leq x \leq \frac{ \pi }{8} + \frac{ \pi n}{2} ,$ $n$ ∈ $Z$