Solving trig equations (1 Viewer)

[kazemagic]

This question... For the final answers, I got thesinX(5cosX-1)=0 , sinX= -3/4

The problem is that the answers in this book for this question only took in consideration of (5cosX-1)=0 and sinX= -3/4

What happened to the value of sinX=0, why isn't it in the answers? Or did I solve it incorrectly?

 

[deswa1]

I think you solved it incorrectly. I just did it quickly (sorry, don't have time to post full working), and I got (5cosx-1)(4+3/sinx)=0 which gives the two answers above. I think you might have made a mistake when you had the sinx on the denominator and multiplied through by sinx and then made an algebraic mistake... Hope that helps

 

[kazemagic]

Kk got it. Thanks! Yep did an algebra mistake. I skipped one step and this is what I got =.= Such a fail

 

[Drongoski]

If you convert the cot's and cosec's in terms of equivalent sin & cos and multipying thru'out by sin^x, you end up with (5cosx - 1)(4sinx + 3) = 0 after factorising.

But you must note that in the original sin x must not = 0 or else cosec x and cot x are undefined, should you somehow end up with sin x as a factor in some simplified eqn LHS = 0.

 

[Sanjeet]

^yeah what he said. Even if you stuck with your working sinx=0 cannot be a solution as cotx would become undefined

 

[kazemagic]

Yer thanks guys. The problem was after this line of working out:
5cotX=1/sinX

I retardedly got to:
tanX=sinX/5

lol