as cos^2 t < = 1 and tan ^2 a >= 0 so the 1st term is
1 and 2nd term is zero.
So cos^2(π/4(sinx+√2 cos2x)) = 1 and tan^2(x+π/4tan^2x) = 0
So (x+π/4tan^2 x) = n π , x = 0 is a solution and other solutions are
= n π – π/4(this I found by guessing tan ^2x =1 and not rigorously)
cos^2(π/4(sin x+√2cos^2x)) = 1 when x = 2 n π – π/4
So cos^2(π/4(sinx+√2 cos2x)) = 1 and tan^2(x+π/4tan^2x) = 0
So (x+π/4tan^2 x) = n π , x = 0 is a solution and other solutions are
= n π – π/4(this I found by guessing tan ^2x =1 and not rigorously)
cos^2(π/4(sin x+√2cos^2x)) = 1 when x = 2 n π – π/4
hence this is the solution
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