Question #cbef0

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Answer 1 · 2017-03-03T15:13:47

$Cos(2θ) + 34 sin2 θ = 25$

$=>1/sqrt(1^2+34^2)Cos(2θ) + 34/sqrt(1^2+34^2) sin2 θ = 25/sqrt(1^2+34^2)$

Let

$cosalpha=1/sqrt(1^2+34^2)$

$sinalpha=34/sqrt(1^2+34^2)$

and
$cos phi =25/sqrt(1^2+34^2)=> phi=cos^-1(25/sqrt(1^2+34^2)) =0.24pi$

So $tanalpha=34=>alpha = tan^-1(34)=0.49pi$

The given equation becomes

$cos2thetacosalpha+sin2thetasinalpha=coa phi$

$=>cos(2theta-alpha)=cosphi$

$=>cos(2theta-0.49pi)=cos(0.24pi)=cos (2pi-0.24pi)$
so
$=>2theta=(0.24+0.49)pi$
$=>theta=(0..365)pi$

Again

$=>2theta=(2-0.24+0.49)pi$

$=>theta=1.125pi$

Further taking

$cos(2theta-0.49pi)=cos(0.24pi)=cos (2pi+0.24pi)$

$=>2theta=(2+0.24+0.49)pi$

$=>theta=1.365pi$