0 0 votes $\sin x - \cos x > 0$ how to solve ? Calculus + – sumit goyal 1 964 views answer comment Share Follow Print See all 6 Comments 6 6 Comments reply MiNiPanda commented Jan 30, 2018 i edited by MiNiPanda Jan 30, 2018 reply Follow flag @vishal chugh Thanks for correcting me sinx-cosx>0 => tanx>1 => π/4 < x < π/2 and 3π/4 < x < 3π/2 0 0 replyShare vishal chugh commented Jan 30, 2018 reply Follow flag @MiNiPanda I think you did a mistake in calculation as for sinx-cosx>0, tan x > 1. The answer should be $\pi /4$ < x < 3$\pi /4$ 1 1 replyShare Rishabh Gupta 2 commented Jan 30, 2018 reply Follow flag sinx > cosx just draw both the graphs and see where the condition is satisfied. 0 0 replyShare sumit goyal 1 commented Jan 30, 2018 reply Follow flag i have to find interval bro for solving some integral Rishabh Gupta 2 you find what you get 0 0 replyShare Rishabh Gupta 2 commented Jan 30, 2018 reply Follow flag $\frac{\pi}{4}$ to $\frac{5\pi}{4}$ 0 0 replyShare Rishabh Gupta 2 commented Jan 30, 2018 reply Follow flag There will be infinte ranges where this condition will hold. (Ofcourse since both are periodic functions). Within [0, 2pi] we have the range that I commented above. 0 0 replyShare Please log in or register to add a comment.
