Asked by Piwokazi
What's the general solution for 3sin^2x+sinxcosx=2
Answers
Answered by
Mgraph
3sin^2(x)+sin(x)cos(x)=2sin^2(x)+2cos^2(x)
sin^2(x)+sin(x)cos(x)-2cos^2(x)=0
Divide by cos^2(x)
tan^2(x)+tan(x)-2=0
(tan(x)+2)(tan(x)-1)=0
tan(x)=-2 => x=-Arctan(2)+pi*n
tan(x)=1 => x=pi/4+pi*n
sin^2(x)+sin(x)cos(x)-2cos^2(x)=0
Divide by cos^2(x)
tan^2(x)+tan(x)-2=0
(tan(x)+2)(tan(x)-1)=0
tan(x)=-2 => x=-Arctan(2)+pi*n
tan(x)=1 => x=pi/4+pi*n
Answered by
Piwokazi
Thnx hey next culd plz explain the steps step by step thnx onc agn
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