If 2sin theta +3cos theta=2 prove that 3sin theta - 2 cos theta =±3?

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Answer 1 · 2018-02-27T16:19:19

Please see below.

Given #rarr2sinx+3cosx=2#

#rarr2sinx=2-3cosx#

#rarr(2sinx)^2=(2-3cosx)^2#

#rarr4sin^2x=4-6cosx+9cos^2x#

#rarrcancel(4)-4cos^2x=cancel(4)-6cosx+9cos^2x#

#rarr13cos^2x-6cosx=0#

#rarrcosx(13cosx-6)=0#

#rarrcosx=0,6/13#

#rarrx=90°#

Now, #3sinx-2cosx=3sin90°-2cos90°=3#

Answer 2 · 2018-02-27T16:54:31

Given# 2sin theta +3cos theta=2#

Now

#(3sin theta - 2 cos theta)^2#

#=(9sin^2theta-2*3sintheta*2costheta+4cos^2theta#

#=9-9cos^2theta-2*3costheta*2sintheta+4-4sin^2theta#

#=13-((3costheta)^2+2*3costheta*2sintheta+(2sintheta)^2#

#=13-(2sintheta+3costheta)^2#

#=13-2^2=9#

So

#(3sin theta - 2 cos theta)^2=9#

#=>3sin theta - 2 cos theta=pmsqrt9#

#=±3#