The problem is based on simple trignometry identiy so please use your little mind in solving the question
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Can you add more explanation?
More complex, for sure. I find that y = sec ( θ ) is a solution to
1 4 4 y 6 − 1 6 8 y 5 − 5 2 7 y 4 + 5 0 4 y 3 + 2 8 2 y 2 − 3 3 6 y + 2 4 5 = 0 ,
which, fortunately, factors to
( 4 y − 5 ) ( 3 6 y 5 + 3 y 4 − 1 2 8 y 3 − 3 4 2 + 2 8 y − 4 9 ) = 0 ,
which gives y = sec ( θ ) = 4 5 as a solution.
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Probably the last step was difficult to think.....is there any other approach to solve the problem.....using a bit calculus i could just estimate the interwal of the answer
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I'm not sure what approach @Aniket Prakash had in mind that was so simple. Although this is an old problem, I do remember spending quite a bit a time trying to find a less difficult method than establishing and then factoring a degree 6 polynomial.
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just use the trignometric identites the solution become more complex but don't fear latter it become easy the you got answer