# Solve the Following Trigonometric Equation

• May 4th 2009, 03:16 PM
ayoYNr
Solve the Following Trigonometric Equation
Hey, in pre-cal we're going over some of the stuff we learned this semester. Unfortunately for me, I have forgotten a bit of it:

So, here's the problem:
Solve the equation:
sin2x + cos3x = 0 for [0, 2pi)

I wish I could just graph it, but I have to do it algebraically and show all work. Any clues as to how to approach this would be appreciated

Thanks
• May 4th 2009, 04:57 PM
skeeter
Quote:

Originally Posted by ayoYNr
Hey, in pre-cal we're going over some of the stuff we learned this semester. Unfortunately for me, I have forgotten a bit of it:

So, here's the problem:
Solve the equation:
sin2x + cos3x = 0 for [0, 2pi)

I wish I could just graph it, but I have to do it algebraically and show all work. Any clues as to how to approach this would be appreciated

Thanks

first task ... you'll need to expand $\displaystyle \sin(2x)$ and $\displaystyle \cos(3x)$

$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$ ... that's done.

$\displaystyle \cos(3x) = \cos(2x+x) = \cos(2x)\cos{x} - \sin(2x)\sin{x} =$ ... keep going
• May 7th 2009, 05:36 AM
ayoYNr
Ok, so now I have:

2sinxcosx + cos(2x)cosx - sin(2x)sinx

Where do I go from here?
• May 10th 2009, 11:57 AM
ayoYNr
(Sorry, ignore this post)
• May 10th 2009, 12:43 PM
skeeter
Quote:

Originally Posted by ayoYNr
Ok, so now I have:

2sinxcosx + cos(2x)cosx - sin(2x)sinx

Where do I go from here?

$\displaystyle 2\sin{x}\cos{x} + (1 - 2\sin^2{x})\cos{x} - 2\sin^2{x}\cos{x} = 0$

$\displaystyle \cos{x}(2\sin{x} + 1 - 2\sin^2{x} - 2\sin^2{x}) = 0$

$\displaystyle \cos{x}(1 + 2\sin{x} - 4\sin^2{x}) = 0$

$\displaystyle \cos{x} = 0$ ... should be straightforward to solve for x

$\displaystyle 1 + 2\sin{x} - 4\sin^2{x} = 0$ ... you'll need to use the quadratic formula to solve for $\displaystyle \sin{x}$