Question

Попалось на егэ, с1. нужна экспертов. 2cos²x + 2√2cos(π/2 - x) + 1 = 0

Answer 1

$2cos^{2}x+2 \sqrt{2}*sinx+1=0$
$2*(1-sin^{2}x)+2 \sqrt{2}*sinx+1=0$
$2-2sin^{2}x+2 \sqrt{2}*sinx+1=0$
$-2sin^{2}x+2 \sqrt{2}*sinx+3=0$
$2sin^{2}x-2 \sqrt{2}*sinx-3=0$

Замена: sinx=t∈[-1;1]

$2t^{2}-2 \sqrt{2}*t-3=0, D=8+4*3*2=32$
$t_{1}= \frac{2 \sqrt{2}-\sqrt{32}}{4}=\frac{2 \sqrt{2}-4\sqrt{2}}{4}=\frac{\sqrt{2}-2\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}$
$t_{2}= \frac{2 \sqrt{2}+\sqrt{32}}{4}=\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\ \textgreater \ 1$ - посторонний корень

Вернемся к замене:
$sinx=-\frac{\sqrt{2}}{2}$
$x=-\frac{ \pi }{4}+2 \pi k$, k∈Z
$x=-\frac{3 \pi }{4}+2 \pi k$, k∈Z