Solve cos^2 x +cos^2 2x +cos^2 3x=1?

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Answer 1 · 2018-06-21T03:10:17

Please see below.

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$cos^2x+cos^2(2x)+cos^2(3x)=1$

$(1+cos2x)/2+(1+cos4x)/2+cos^2(3x)=1$

$1+cos2x+1+cos4x+2cos^2(3x)=2$

$2+cos2x+cos4x+2cos^2(3x)=2$

$cos2x+cos4x+2cos^2(3x)=0$

$2cos((4x+2x)/2)cos((4x-2x)/2)+2cos^2(3x)=0$

$cos3xcosx+cos^2(3x)=0$

$cos3x(cos3x+cosx)=0$

$cos3x=0, :. 3x=pi/2+-2kpi,(3pi)/2+-2kpi, :. x=pi/6+-2/3kpi,pi/2+-2/3kpi$

$cos3x+cosx=0$

$cos3x=-cosx$

$3x=pi+x+-2kpi, :. 2x=(1+-2k)pi, :. x=pi/2(1+-2k)$

$3x=pi-x+-2kpi, :. 4x=(1+-2k)pi, :. x=pi/4(1+-2k)$