savianp
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# Find all solutions of the equation in the interval [0, 2pi]

$sec^2(y)+6*tan(y)+4=0$ first we have to change this $sec^2(y)$ we know this equation $sin^2(y)+cos^2(y)=1$ then, let's divide all members by $cos^2(y)$ $\frac{sin^2(y)}{cos^2(y)}+\frac{cos^2(y)}{cos^2(y)}=\frac{1}{cos^2(y)}$ then got... $tan^2(y)+1=sec^2(y)$ now we can replace it at original equation $tan^2(y)+1+6*tan(y)+4=0$ $tan^2(y)+6*tan(y)+5=0$ now we have to make another substitution... $tan(y)=x$ so $x^2+6*x+5=0$ then we found $x_1=-5~~and~~x_2=-1$ now we have to back to tan(y) $tan(y)=-1$ $y=arctan(-1)$ $y=\frac{3\pi}{4}~~and~~y=\frac{7\pi}{4}$ if you have a scientific calculator you'll find out this value to $tan(y)=-5$ in degrees $y\approx281.31\º~~and~~y\approx101.31\º$