How do you solve #sin^2x-5cosx=5# and find all solutions in the interval #[0,2pi)#?

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Answer 1 · 2018-07-27T19:02:51

#x=\pi#

Given equation:

#\sin^2x-5\cos x=5#

#1-\cos^2x-5\cos x=5#

#\cos^2x+5\cos x+4=0#

#\cos^2x+\cos x+4\cos x+4=0#

#\cos x(\cos x+1)+4(\cos x+1)=0#

#(\cos x+1)(\cos x+4)=0#

#\cos x+1=0\ \quad (\because \cosx\ne -4)#

#cos x=-1#

#\cos x=\cos \pi#

#x=(2k+1)\pi#

Where, #k# is any integer i.e. #k=0, \pm1, \pm2, \pm3, \ldots#

But #x\in [0, 2\pi)# hence setting #k=0# we get the solution

#x=\pi#

Answer 2 · 2018-07-27T19:04:51

#x=pi#

Here,

#sin^2x-5cosx=5# , #where, x in [0,2pi)#

#=>1-cos^2x-5cosx=5#

#=>cos^2x+5cosx+4=0#

#=>cos^2x+4cosx+cosx+4=0#

#=>cosx(cosx+4)+1(cosx+4)=0#

#=>(cosx+1)(cosx+4)=0#

#=>cosx=-1 or cosx=-4 !in [-1,1]#

#;.cosx=-1# #and x in [0,2pi)#

So , #x=pi#