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Optimization: Trig functions

8(sinX)^-2.cosX - 4(cosX)^-2.sinX = 0

is my derivative


How do I solve for X easily?
Reply 1
Original post by chrislpp
8(sinX)^-2.cosX - 4(cosX)^-2.sinX = 0

is my derivative


How do I solve for X easily?


8cosxsin2x4sinxcos2x=0    2cosxsin2x=sinxcos2x \displaystyle \frac{8cosx}{sin^2x} - \frac{4sinx}{cos^2x} = 0 \implies \frac{2cosx}{sin^2x} = \frac{sinx}{cos^2x}

Multiply both sides by sin2xcos2x sin^2xcos^2x , then divide both sides by cos3x cos^3x

You will get tan3x=2    tanx=23 tan^3 x = 2 \implies tanx = \sqrt[3]2 Solve it to get the solutions.

Sorry for not replying to your previous thread, i did saw that you quoted me about this question but i was busy so couldn't help.
Reply 2
Add 4(cosX)^-2.sinX to both sides (to get an equality), get rid of the fractions, divide one side by the other.
Reply 3
Original post by raheem94
.. Solve it to get the solutions. This is a bit close to a full solution if you ask me...
Reply 4
Original post by DFranklin
This is a bit close to a full solution if you ask me...


I understand it is close to a full solution. Actually he asked me this question on another thread, but i was busy so can't reply neither did anyone else helped him with it, so i thought that he has been stuck on it for a long time now and as no one was helping him so i decided to give him enough detail to finish off the question.

Sorry for it though.
Reply 5
Original post by raheem94
I understand it is close to a full solution. Actually he asked me this question on another thread, but i was busy so can't reply neither did anyone else helped him with it, so i thought that he has been stuck on it for a long time now and as no one was helping him so i decided to give him enough detail to finish off the question.

Sorry for it though.
No worries.

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