Question #6dd1a

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Answer 1 · 2017-03-03T13:27:16

$sin2x=4cos^2x-6sin^2x$

$=>6sin^2x+2sinxcosx-4cos^2x=0$

$=>3sin^2x+sinxcosx-2cos^2x=0$

$=>(3sin^2x)/cos^2x+(sinxcosx)/cos^2x-(2cos^2x)/cos^2x=0$

$=>3tan^2x+tanx-2=0$

$=>3tan^2x+3tanx-2tanx-2=0$

$=>3tanx(tanx+1)-2(tanx+1)=0$

$=>(tanx+1)(3tanx-2)=0$

When $tanx+1=0$

$=>tanx=-1=tan(pi-pi/4)=tan(2pi-pi/4)$

So

$x=(3pi)/4 and (7pi)/4$

$=>x=135^@ and 315^@$

Again

$3tanx -2 = 0$

$=>tanx=2/3=tan(tan^-1(2/3))=tan33.7^@$

So

$x= 33.7^@ and (180+33.7)=213.7^@$