How can this be solved?

Question

#4sec^2(x) - 11sec(x) - 3#

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Answer 1 · 2018-03-02T09:12:41

#x~~2pin-1.23,2pin+1.23#, #ninZZ#

We want to solve

#4sec^2(x)-11sec(x)-3=0#

Let #u=sec(x)#

#4u^2-11u-3=0#

Solve this quadratic equation

#4u^2-11u-3=0#

#=>(u-3)(4u+1)=0#

#=>u=-1/4,3#

#=>sec(x)=-1/4,3#

The range of secant is #(-oo,-1]uu[1,oo)#, thus we seek

#sec(x)=3=>cos(x)=1/3=>x~~2pin-1.23,2pin+1.23#