Question

Question #cbef0

Answer 1

`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`