Question

plzz anser this question

Answer 1

incorrect answer ???????????

Answer 2

I HAVE TAKEN THETA =A BECAUSE OF ABSENCE OF THETA SYMBOL
2sin^2A+5cosA=4

2sin^2A+5cosA-4=0

2(1-cos^2A)+5cosA-4=0

2-2cos^2A+5cos^A-4=0

-2-2cos^2A+5cosA=0

Divide the above eq. by 1,we get

2+2cos^2A-5cosA=0

We can also write it as

2cos^A-5cosA+2=0

Now by splitting the middle term, we get

2cos^A-4cosA-cosA+2=0

2cosA(cosA-2)-1(cosA-2)=0

(cosA-2) (2cosA-1)=0

Firstly put (cosA-2)=0 | Secondly put (2cosA-1)=0
cosA-2=0 | 2cosA-1=0
cosA=2 | 2cosA=1 , cosA=1/2

Therefore , the value of cosA can be either 2 or 1/2
Hope this will help you
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