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vaibhav206
@vaibhav206
June 2022
2
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plzz anser this question
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palkaushal
Incorrect answer ???????????
0 votes
Thanks 1
vaibhav206
it is not correct we must prove cosa=1/2
Tejasv098
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
Comment me if have any doubt
0 votes
Thanks 1
vaibhav206
thanks
Tejasv098
np
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Answers & Comments
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
Comment me if have any doubt