Trigonometry questions (1 Viewer)

[hsc2013]

Could someone please show me the working out for these questions?

1.sin3x =sin2x
2. sin2x=tanx
3. tan(a - pi/3) = -3
4. sin 4x-sin2x=0
5. cos3x = cosx
6. 3tan2x = 2tanx

 

[deswa1]

Is the domain restricted or do you want a general answer? The reason is I don't want to spend half an hour typing something in latex (I'm very unfluent) if that's not the answer you're looking for

 

[SpiralFlex]

Is the domain restricted or do you want a general answer? The reason is I don't want to spend half an hour typing something in latex (I'm very unfluent) if that's not the answer you're looking for

I am sure I saw a program that converts handwriting into LaTeX.

Now you know how I feel.

 

[hsc2013]

Oh sorry, I just want the general answer

 

[Carrotsticks]

I'm pretty sure I saw Question 1 in another thread somewhere.

 

[mnmaa]

Theres my working for question 1 i cant be bothered writing up the rest at the moment tbh

 

[mnmaa]

actually nvm
for the second question:
sin2x=tanx
2sinxcosx=sinx/cosx
2cosx=1/cosx just divided both sides by sinx
2cos^2x=1
let cosx=u
therefore
2u^2=1
u=plus and minus(√2/2)
cosx=plus and minus(√2/2)
solve for x

 

[deswa1]

actually nvm
for the second question:
sin2x=tanx
2sinxcosx=sinx/cosx
2cosx=1/cosx just divided both sides by sinx
2cos^2x=1
let cosx=u
therefore
2u^2=1
u=plus and minus(√2/2)
cosx=plus and minus(√2/2)
solve for x

You can't just divide both sides by sinx because you remove the solution sinx=0. As a general rule, you shouldn't divide by a function of x unless you know that x=0 isn't a solution.

A neater way to do the first one is to do:
sin3x-3in2x=0
2cos(5x/2)sin(x/2)=0 (factorising)
cos(5x/2)=0 or sin(x/2)=0
x=(pi+2kpi)/5 or x=2kpi for integers k

 

[Carrotsticks]

You can't just divide both sides by sinx because you remove the solution sinx=0. As a general rule, you shouldn't divide by a function of x unless you know that x=0 isn't a solution.

^^

 

[hayabusaboston]

Could someone please show me the working out for these questions?

1.sin3x =sin2x
2. sin2x=tanx
3. tan(a - pi/3) = -3
4. sin 4x-sin2x=0
5. cos3x = cosx
6. 3tan2x = 2tanx

this is the same guy, capper

 

[mnmaa]

You can't just divide both sides by sinx because you remove the solution sinx=0. As a general rule, you shouldn't divide by a function of x unless you know that x=0 isn't a solution.

A neater way to do the first one is to do:
sin3x-3in2x=0
2cos(5x/2)sin(x/2)=0 (factorising)
cos(5x/2)=0 or sin(x/2)=0
x=(pi+2kpi)/5 or x=2kpi for integers k

My bad , post year 12 :/ lol

 

[Carrotsticks]

this is the same guy, capper

Hahahaha you're back on green bar!

Try to go for a record for the most green/red bar transitions in a day.

btw I can't PM you. It says you blocked the function.

 

[bleakarcher]

Dude, ive been stuck on one bar forever!

EDIT:

Im on two now!

 

[RivalryofTroll]

Dude, ive been stuck on one bar forever!

EDIT:

Im on two now!

I was stuck on 1 bar for quite awhile as well....