How do you find the solutions of this? #3cosxsinx-sin^2x=1#

#3cosxsinx-sin^2x=1#

1 Answer
Sunayan S
Feb 11, 2018

See the answer below...

Explanation:

#3cosxsinx-sin^2x=1#
#=>(3cosxsinx)/cos^2x-sin^2x/cos^2x=1/cos^2x#
#=>3tanx-tan^2x=sec^2x#
#=>3tanx-tan^2x=1+tan^2x#
#=>2tan^2x-3tanx+1=0#
#=>2tan^2x-2tanx-tanx+1=0#
#=>2tanx(tanx-1)-1(tanx-1)=0#
#=>(2tanx-1)(tanx-1)=0#

Either,

#(2tanx-1)=0#
#=>tanx=1/2=tanalpha#
#=>color(red)(ul(bar(|color(green)(x=npi+alpha)|##" "#where #" "alpha=tan^-1(1/2)" ; "n in I#

or,

#tanx-1=0#
#=>tanx=1#
#=>tanx=tan(pi/4)#
#=>color(red)(ul(bar(|color(green)(x=npi+(pi/4))|))" "[n in I]#

Hope it helps...
Thank you...

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