Maths!

Hi, am stuck with the question below. I thought I was working it out correctly, but 2 of the values I find are supposdly wrong. Could someone take a look and point out my mistake?! Thanks

Q. Solve 3cot2x + cotx = 1 for values of x from 0-360 degrees inclusive

My method:

3[(1-cot^2x)/2cotx] + cot x = 1

3 - 3cot^2x + 2cot^2x = 2cotx

cot^2 + 2ctox - 3 = 0
(cotx - 1)(cot x +3) = 0

So either cot x - 1 = 0 which gives x = 45 & 225 (these are correct)

OR cot x + 3 = 0 which implies x=-18.4 (not in range); 161 & 341.6

However, the answers given are 45, 225, 121 & 301!

HELPPPP!!!!!!!!!!!

Reply 1

JamesF

Re-write in terms of tan x, then use the double angle formula to write tan2x in terms of tan x. Then you get a quadratic in tan x.
I get tan x = 1 , -5/3

So x = 45, 225, 120.96, 300.96

Reply 2

username9816

JamesF

Re-write in terms of tan x, then use the double angle formula to write tan2x in terms of tan x. Then you get a quadratic in tan x.
I get tan x = 1 , -5/3

So x = 45, 225, 120.96, 300.96

What's cot? :confused:

Reply 3

m:)ckel

bono

What's cot? :confused:

cotx = 1/tanx
secx = 1/cosx
cosecx = 1/sinx

you learn this in P2, with loads more trig identities

Reply 4

kikzen

mockel

cotx = 1/tanx
secx = 1/cosx
cosecx = 1/sinx

you learn this in P2, with loads more trig identities

not to be confused with tanx^-1, cos^-1 and sin^-1 (arctan, arccos, arcsin)

Reply 5

makesomenoise

bono

What's cot? :confused:

Cot, or the cotangent function, the reciprocal of the tangent function. So cotX = 1/tanX = cosX/sinX (since tanX = sinX/cosX).

You'll also learn of the secant and cosecant functions. Secant (sec) is the reciprocal of cos, and cosecant (cosec or csc) is the reciprocal of sin.

secX = 1/cosX
cosecX = 1/sinX
cotX = 1/tanX = cosX/sinX

Reply 6

john !!

I thought sin^-1x would be the same as 1/sinx...

Reply 7

Nylex

mik1a

I thought sin^-1x would be the same as 1/sinx...

You usually refer to sin^-1 x as the inverse sine (same with the other trig/hyperbolic functions). If you mean 1/sin x, better to write either cosec x or (sin x)^-1 cos that way, you won't get confused.

Reply 8

makesomenoise

mik1a

I thought sin^-1x would be the same as 1/sinx...

That confused me too at first. But they have different curves. For some reason.
I just think of it as: if f(X) = sinX, arcsinX = f^-1(X) and cosecX = (f(X))^-1, I doubt that helps but I hope it does...