Another Trigonometric Equation

[TrueStar]

Homework Statement

Solve for t

sin3tcost+cos3tsint=-1/2

Homework Equations

Sum and Difference Identities

The Attempt at a Solution

I saw this problem the the sum identity sin($\alpha$+$$\beta$$) looked like the way to begin solving for t.

This condenses to sin(3t+t)= -1/2 which is sin(4t)= -1/2

I let 4t=x and now have sin x= -1/2

The answers for this are 7$\pi$/6+2$\pi$n and 11$\pi$+2$\pi$n.

This means that 4t equals both 7$\pi$/6+2$\pi$n and 11$\pi$+2$\pi$n. I need to divide the four out to get t by itself.

The general solutions (I think) are t=7$\pi$/24+$\pi$n/2 and 11$\pi$/6+$\pi$n/2.

Did I miss anything or does this look correct?

Thanks!

 

[Mark44]

Homework Statement

Solve for t

sin3tcost+cos3tsint=-1/2

Homework Equations

Sum and Difference Identities

The Attempt at a Solution

I saw this problem the the sum identity sin($\alpha$+$$\beta$$) looked like the way to begin solving for t.

This condenses to sin(3t+t)= -1/2 which is sin(4t)= -1/2

I let 4t=x and now have sin x= -1/2

In the next couple of lines, the 2nd expression should be 11$\pi$/6 +2$\pi$n, which it appears you caught later.

The answers for this are 7$\pi$/6+2$\pi$n and 11$\pi$+2$\pi$n.

This means that 4t equals both 7$\pi$/6+2$\pi$n and 11$\pi$+2$\pi$n. I need to divide the four out to get t by itself.

In the next line, you got the "divided by 6" back in the second expression, but forgot to divide by 4.

The general solutions (I think) are t=7$\pi$/24+$\pi$n/2 and 11$\pi$/6+$\pi$n/2.

Make the above t=7$\pi$/24+$\pi$n/2 and 11$\pi$/24+$\pi$n/2.

Did I miss anything or does this look correct?

Thanks!

 

[TrueStar]

You're right. I had that written on my paper, but it's a typo here on the board. Thank you for reading my post!