Solve 4sin(2t)+ 4 = 8cos(t) - 4sin(t) for all real solutions?

[mauro125]

Solve 4sin(2t)- 4 = 8cos(t) - 4sin(t) for all real solutions?

Hello, having a little trouble with my trig class, wondering if somebody could help me, so far what i think I should do is move everything to either side, then set them equal to zero

4sin(2t)- 4 - 8cos(t) + 4sin(t) = 0

4sin(2t) + 4 = 0
sin(2t) = -1 use double angle identity?

-8cos(t) = 0
cos(t) = 0
cos is in range (0,pi), also cos is + in QI and QIV so the only possible answer where cos is 0 would be pi/2?

4sin(t) = 0
sin(t) = 0
sin is in range [-pi/2, pi/2], sin is positive in QI and QII so the only possible answer would be 0?



any help would be greatly appreciated thanks

edit:font this color old info

 

[HallsofIvy]

The problem is NOT trig, it is basic algebra. When you have an equation that involves three that, added together, equal to 0, you cannot set each of them equal to 0! There has to be one value of t that, put into the equation, make the left side equal to 0, not a different t for each part. (It is true that "if ab= 0 then either a= 0 or b= 0 or both." It is NOT true that "if a+ b= 0 then a= 0 or b=0 or both.") By the way cos(0)=1, not 0.

 

[mauro125]

thanks for the help, pretty new at this so i'm just trying things out, one does not learn to walk without falling first..
should i factor out a 4 and see where that leads me?

 

[stapel]

...so far what i think I should do is move everything to either side, then set them equal to zero

4sin(2t)+ 4 - 8cos(t) + 4sin(t) = 0

What was the original exercise?

Assuming you did the moving around correctly, it might be useful to apply a double-angle formula to the sine of 2t, in order to get everything into sines and cosines of "just t".

. . . . .8sin(t)cos(t) - 8cos(t) + 4sin(t) + 4 = 0

Dividing through by 4 will simplify things somewhat:

. . . . .2sin(t)cos(t) - 2cos(t) + sin(t) + 1 = 0

Maybe try factoring in some manner at this point...?

Note: We're still just working with algebra. Once you get this factored, then you can solve for two trig equations, each of which you can solve for t-values. It's only at that final solving that you'll be bringing trig into the mix.

 

[mauro125]

What was the original exercise?

Assuming you did the moving around correctly, it might be useful to apply a double-angle formula to the sine of 2t, in order to get everything into sines and cosines of "just t".

. . . . .8sin(t)cos(t) - 8cos(t) + 4sin(t) + 4 = 0

Dividing through by 4 will simplify things somewhat:

. . . . .2sin(t)cos(t) - 2cos(t) + sin(t) + 1 = 0

Maybe try factoring in some manner at this point...?

Note: We're still just working with algebra. Once you get this factored, then you can solve for two trig equations, each of which you can solve for t-values. It's only at that final solving that you'll be bringing trig into the mix.

thanks for the tips, this is original probelm

Solve 4sin(2t)-4 = 8cos(t)-4sin(t) for all real solutions of t. Write your answer(s) in a solution set.

i had copied wrong, i updated first post


the reason I tried to split it up was because of a similar example from proff, pretty much what you just told me thanks!