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C4 trig Q

I'm really lost on this one

solve cotx.cos2x=0
interval 0<x<180 (strict inequality not 'or equals to')

MS is as follows:
cotx=0 or cos2x=90

2x=90, 270

x= 90, 45, 135

the main thing i dont understand is why:
cotx=0
so 1/tanx=0
so tanx=0
so x=180 (out of x range)
is not valid?
Original post by hi-zen-berg
I'm really lost on this one

solve cotx.cos2x=0
interval 0<x<180 (strict inequality not 'or equals to':wink:

MS is as follows:
cotx=0 or cos2x=90

2x=90, 270

x= 90, 45, 135

the main thing i dont understand is why:
cotx=0
so 1/tanx=0
so tanx=0
so x=180 (out of x range)
is not valid?

1/tanx = 0 implies that tanx = infinity, not 0.

If tanx = 0 then 1/0 would leave you with.. a lot of problems :tongue: (your calculator would not be able to find it, for one!)

So tanx is not 0. 1/tanx is.

So we know that cotx = 0 when tanx = infinity, and that is the case only at pi/2 between 0 and pi, so pi/2 = 90 is the solution from that.
(edited 7 years ago)
Original post by hi-zen-berg
I'm really lost on this one

solve cotx.cos2x=0
interval 0<x<180 (strict inequality not 'or equals to':wink:

MS is as follows:
cotx=0 or cos2x=90

2x=90, 270

x= 90, 45, 135

the main thing i dont understand is why:
cotx=0
so 1/tanx=0
so tanx=0
so x=180 (out of x range)
is not valid?


You ifhave cotx.cos2x=0

Hence either cotx or cos2x = 0

If cotx = 0, then the reciprocal of tanx, i.e. cosx/sinx=0, hence cosx must be 0. This occurs when x is 90 or 270, but seeing as the latter is outside the range we ignore it.

If cos2x=0, 2x = 90 or 270, hence x is either 45 or 135,giving us 3 solns in total.

Referring back to ur question, if 1/tanx =0, then if x was 180, tan x would be 0, so 1/tanx would be 1/0. Hence the answer would be undefined/infinity rather than 0. Also the markscheme already says that the solution is not defined for 180 as its a strict inequality so we ignore that possibility anyway.

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