Question

5/cos^2 x +7/sin(5p/2-x) +2=0

Answer 1

5/cos²x+7/cosx+2=0
cosx≠0
2cos²x+7cosx+5=0
cosx=a
2a²+7a+5=0
D=49-40=9
a1=(-7-3)/4=-2,5⇒cosx=-2,5<-1 нет решения
a2=(-7+3)/4==-1⇒cosx=-1⇒x=π+2πn,n∈z

Answer 2

$\frac{5}{cos^2x} + \frac{7}{sin(\frac{5\pi }{2}-x)} +2=0\\\\sin(\frac{5\pi }{2}-x)=cosx\; \; \Rightarrow \; \; \frac{5}{cos^2x} + \frac{7}{cosx} +2=0\, |\cdot cos^2x\ne 0\\\\ \frac{5+7cosx+2cos^2x}{cos^2x} =0\; ,\; cosx\ne 0\\\\2cos^2x+7cosx+5=0\\\\D=9\; ,\; (cosx)_1=-\frac{5}{2}\; \; net\; reshenij\\\\(cosx)_2=-1\; ,\; x= \pi +2\pi n,\; n\in Z$