What are the smallest four positive solutions for 9cos(2t)=9cos2(t)-4?

[nickb145]

Homework Statement

9cos(2t)=9cos²(t)-4

for all the smallest four positive solutions

Homework Equations

The Attempt at a Solution

I've factored it and pulled out 9cos(t) and made that = u
so i have u²-2u-4
That factors to 3.23606 and -1.236

Next i added 9cos(t) back in. 9cos(t)=3.236 and 9cos(t)=-1.236
then solve for the values for the inverse cos^-1(3.236/9) and cos^-1(-1.236/9)



am i on the right track? I'm not entirely sure if I'm doing it correctly

 

[Mark44]

Homework Statement

9cos(2t)=9cos²(t)-4

for all the smallest four positive solutions

Homework Equations

The Attempt at a Solution

I've factored it and pulled out 9cos(t) and made that = u
so i have u²-2u-4
That factors to 3.23606 and -1.236

Next i added 9cos(t) back in. 9cos(t)=3.236 and 9cos(t)=-1.236
then solve for the values for the inverse cos^-1(3.236/9) and cos^-1(-1.236/9)



am i on the right track?

No. You are thinking the equation is quadratic in form - it isn't. Use the double-angle identity to rewrite cos(2t) in a different form.

I'm not entirely sure if I'm doing it correctly

 

[nickb145]

No. You are thinking the equation is quadratic in form - it isn't. Use the double-angle identity to rewrite cos(2t) in a different form.

ak ok now i have 8cos²(t)-9sin²(t)+4

is this better?

 

[Dick]

ak ok now i have 8cos²(t)-9sin²(t)+4

is this better?

How did you get that??

 

[nickb145]

How did you get that??

ok so i used the double angle forumla cos(2t)=cos²(t)-sin²(t)
9(cos²(t)-sin²(t))=cos²(t)-4
distributed the 9 and 9cos²(t)-9sin²(t))=cos²(t)-4

then subtracted cos² and added 4. 8cos2(t)-9sin2(t)+4

I'm not doing something right...
I'm thinking i need the half angle formula cos²=1+cos(2u)/2

 

[nickb145]

whoops messed something up here

 

[Dick]

ok so i used the double angle forumla cos(2t)=cos²(t)-sin²(t)
9(cos²(t)-sin²(t))=cos²(t)-4
distributed the 9 and 9cos²(t)-9sin²(t))=cos²(t)-4

then subtracted cos² and added 4. 8cos2(t)-9sin2(t)+4

I'm not doing something right...
I'm thinking i need the half angle formula cos²=1+cos(2u)/2

The original equation you posted was 9cos(2t)=9cos²(t)-4. What happened to the second 9? Or was that a typo?

 

[nickb145]

The original equation you posted was 9cos(2t)=9cos²(t)-4. What happened to the second 9? Or was that a typo?

typo. the 9cos²(t) canceled out now i have -9sin²=-4 -> sin²=4/9

 

[Dick]

typo. the 9cos²(t) canceled out now i have -9sin²=-4 -> sin²=4/9

That's better.

 

[nickb145]

That's better.

now I've squarerooted it and now have sin(t)=2/3 now to arcsin(2/3)=.7297. That is one solution

( i think)

then i think i subtract 2pi. 2pi-.7297=5.5534 (second solution i think). I add and subtract pi to .7297 because 2pi would make it too big

so pi-.7297=2.4118
pi+.72972=3.871

 

[Dick]

now I've squarerooted it and now have sin(t)=2/3 now to arcsin(2/3)=.7297. That is one solution

( i think)

then i think i subtract 2pi. 2pi-.7297=5.5534 (second solution i think). I add and subtract pi to .7297 because 2pi would make it too big

so pi-.7297=2.4118
pi+.72972=3.871

Sounds about right. But your logic is a little fuzzy. sin(t)=(-2/3) is also a solution. Graph y=sin(t), y=2/3 and y=(-2/3) in the range [0,2*pi] to make sure you understand why all of those work.