that was through simple factorization
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A*(a*)
Rutzaba
the one in wch its cos theta
A*(a*)
i did ii as well but my answer didn't match the ms one.
actually we have to take square roots of our answers in (i), taking square root of one the complex root doesn't match from one in ms
actually we have to take square roots of our answers in (i), taking square root of one the complex root doesn't match from one in ms
Rutzaba
no im talking abt the questuion u asked me... is ka frst part as in is se just hle jo hai 8 ii a
A*(a*)
use the results from part (i) to rewrite the equation in terms of cos(theta-alpha) = constant and then solved it. but couldn't do the second one
Rutzaba
cos(theta-alpha) = constant agar aese kr ke agya to part b bhi ajaye ga
consider theta minus alpha to be x
cos 1/2 x wud be cos 1/2 (theta minus alpha) =3
consider theta minus alpha to be x
cos 1/2 x wud be cos 1/2 (theta minus alpha) =3
Rutzaba
the ans is wrong then