Trig

[stiff solar]

Sorry amendment..

Confused on question e here. The textbook gives only two solutions for this answer, which is 7π/6 and 11π/6

However, I ended up with four solutions. Can someone explain why the first two don't count?

[rough elm]

@stiff solar

brotha

squaring adds miscellaneous roots

eg x=-1
but x^2=1 adds x=1 as a root

solve it this way

2 cosx=-cotx

cotx=1/tanx=1/sinx/cosx multiply numerator and denominator by cos x

2 cos x= - cos x/ sin x

cancel cos assuming it is not zero

2=-1/ sinx

multiply by sinx/2 both sides

gives

sinx=-1/2

now see sin x is -ve

so it must be in 3rd/4th quadrant as sin x = sin (pi-x)

sin pi/6=1/2

so after adding pi gives pi+pi/6=7pi/6

subtracting from 2pi gives 2pi-pi/6=11pi/6

hope it helps

[stiff solar]

Thanks for this. Could you elaborate more on what you mean by squaring adds miscellaneous roots?

[rough elm]

see @stiff solar

let me take it in the sense of trig

let sin x = -1/2

then sin^2(x)=1/4

then if you root it

sinx=1/2 is an additional root

[rough elm]

in the second image

you squared it

check all the roots you got

in 2nd img left side last line option 1 and 2 do not satisfy

[stiff solar]

Ah thank you. Yes I see.

[rough elm]

any thing else?

[stiff solar]

That's it. Thanks!

[rough elm]

ok then

[rough elm]

wlcm

close this

[stiff solar]

.close