Question #9ca46

1 Answer
Sunayan S
Oct 30, 2017

See the answer below...

Explanation:

#(sin2x+sin2x+sin2x)^2+(cosx+cosx+cosx)^2=1#
#=>(3cdot sin2x)^2+(3cdotcosx)^2=1#
#=>(3cdot2cdot sinxcdotcosx)^2+(3cdotcosx)^2=1#
#=>36cdot sin^2x cdotcos^2x+9cdotcos^2x=1#
#=>36cdot (1-cos^2x) cdotcos^2x+9cdotcos^2x=1#
[AS PER IDENTITY]
#=>36cos^2x-36cos^4x+9cdotcos^2x=1#
#=>36cos^4x-45cos^2x+1=0#

APPLY SREEDHAR ACHARYA FORMULA:-

#cos^2x=(-(-45)+-sqrt((-45)^2-4cdot36cdot1))/(2cdot36#
#=>cos^2x=(15+sqrt209)/24 or, (15-sqrt209)/24#
#=>cosx=sqrt((15+sqrt209)/24)=1.107 or,sqrt((15-sqrt209)/24)= 0.15#

BY APPROXIMATION

The first value #(1.107)# is negligible because we know #-1<=cosx<=1#

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