Find solutions in decimal form rounding to the nearest thousandth for #sec^2x+4tanx=2# in the interval [0,2pi) ?

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#sec^2x+4tanx=2#

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Answer 1 · 2017-11-14T12:30:35

#x=13.283°,193.283°, 103.283°,283.283°#in the interval #[pi,2pi]#

#sec^2x+4tanx=2#

or, #1+tan^2x+4tanx=2#

or, #tan^2x+4tanx-1=0#

Comparing with #ax^2+bx+c=0# we have #a=1, b=4, c=-1#

#tanx=(-b+-sqrt(b^2-4ac))/(2*a)#

#=(-4+-sqrt(4^2-4*1*(-1)))/(2*1)#

#=(-4+-sqrt(20))/2=(-4+-2*sqrt(5))/2#

Taking +ve sign

#tanx=(2sqrt(5)-4)/2=sqrt(5)-2#

#x=tan^-1(sqrt(5)-2)=13.283°#

Since #tanx=tan(180°+x)#

#tan13.283°=tan(180°+13.283°)=tan193.283°#

#x=13.283°,193.283°#

Taking -ve sign

#tanx=(-2sqrt(5)-4)/2=-(sqrt(5)+2)#

#x=tan^-1(-(sqrt(5)+2))=103.283°#

#tan103.283°=tan(180°+103.283°)=tan283.283°#

#x=103.283°,283.283°#

So, #x=13.283°,193.283°, 103.283°,283.283°#in the interval #[pi,2pi]#.