# Help with 2 trigonometic equations?

• May 2nd 2009, 10:08 PM
chrozer
Help with 2 trigonometic equations?
Quote:

Solve the equation $\sin {2x} + \cos {3x} = 0$ for $[0, 2\pi)$.
I know most of the trigonometric identies, but I'm still stuck on how to solve it.
This is how far I've gotten.
$2\sin{x}\cos{x} + \cos{(x+2x)}$
$2\sin{x}\cos{x} + \cos{x}\cos{2x} - \sin{x}\sin{2x}$
Help?

Quote:

Solve the equation $\cos{2x} - \sin{x} = \frac {1}{2}$ for $[0, 2\pi)$
.
Same problem here. I get stuck and can't solve the problem. Help?
• May 2nd 2009, 10:41 PM
TheEmptySet
Quote:

Originally Posted by chrozer
I know most of the trigonometric identies, but I'm still stuck on how to solve it.
This is how far I've gotten.
$2\sin{x}\cos{x} + \cos{(x+2x)}$
$2\sin{x}\cos{x} + \cos{x}\cos{2x} - \sin{x}\sin{2x}$
Help?

.
Same problem here. I get stuck and can't solve the problem. Help?

Note that $\cos(3x)=4\cos^3(x)-3\cos(x)$
$\sin(2x)=2\sin(x)\cos(x)$

$2\sin(x)\cos(x) + 4\cos^3(x)-3\cos(x)=0$

$\cos(x) [2\sin(x) + 4\cos^2(x)-3]=0$

$\cos(x) [2\sin(x) + 4(1-\sin^{2}(x))-3]=0$

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

$-4\cos(x) [\sin^{2}(x)-\frac{1}{2}\sin(x)-\frac{1}{4}]=0$

$-4\cos(x) [\sin^{2}(x)-\frac{1}{2}\sin(x)+\frac{1}{16}-\frac{1}{16}-\frac{1}{4}]=0$

$-4\cos(x) [\left(\sin(x)-\frac{1}{4} \right)^2-\frac{5}{16}]=0$

I hope this helps

for the 2nd one

$\cos(2x)=1-2\sin^2(x)$

And you will get a quadratic in sine again
• May 2nd 2009, 11:06 PM
pflo
Quote:

Originally Posted by chrozer
$\sin {2x} + \cos {3x} = 0$
$2\sin{x}\cos{x} + \cos{(x+2x)}$
$2\sin{x}\cos{x} + \cos{x}\cos{2x} - \sin{x}\sin{2x}$

Good start. There are probably simplier ways, but I'm a fan of being systematic. I always want to be dealing with the same angle in all my trig functions. I'd use the double-angle identities for both sine and cosine, then simplify and factor so I could use the zero product property to solve:

$2\sin{x}\cos{x}+\cos{x}(1-\sin{x}^2)-\sin{x}(2\sin{x}\cos{x})=0$
$\cos{x}(2\sin{x}+1-\sin{x}^2-2\sin{x}^2)=0$
$\cos{x}(2\sin{x}+1-3\sin{x}^2)=0$
Applying the zero product property...
$\cos{x}=0$
$x=\arccos{x}$
$x=\pm\frac{\pi}{2}+2{\pi}k$
and...
$3\sin{x}^2-s\sin{x}-1=0$
$3\sin{x}+1=0$ and $\sin{x}-1=0$
Solve these (don't forget that you'll get multiple solutions like above with the cosine - then limit them to your interval).

Quote:

Originally Posted by chrozer
$\cos{2x} - \sin{x} = \frac {1}{2}$

Follow the same procedure for this problem. Get all the angles the same in your trig functions, then use the zero product property to solve. With the cosine double-angle identity, use $\cos{2t}=1-2\sin{t}^2$ instead of the other versions so that you'll have something to factor out (the $\sin{x}$) once you get it equal to zero.
• May 3rd 2009, 10:15 AM
chrozer
Quote:

Originally Posted by pflo
Good start. There are probably simplier ways, but I'm a fan of being systematic. I always want to be dealing with the same angle in all my trig functions. I'd use the double-angle identities for both sine and cosine, then simplify and factor so I could use the zero product property to solve:

$2\sin{x}\cos{x}+\cos{x}(1-\sin{x}^2)-\sin{x}(2\sin{x}\cos{x})=0$
$\cos{x}(2\sin{x}+1-\sin{x}^2-2\sin{x}^2)=0$
$\cos{x}(2\sin{x}+1-3\sin{x}^2)=0$
Applying the zero product property...
$\cos{x}=0$
$x=\arccos{x}$
$x=\pm\frac{\pi}{2}+2{\pi}k$
and...
$3\sin{x}^2-s\sin{x}-1=0$
$3\sin{x}+1=0$ and $\sin{x}-1=0$
Solve these (don't forget that you'll get multiple solutions like above with the cosine - then limit them to your interval).

Follow the same procedure for this problem. Get all the angles the same in your trig functions, then use the zero product property to solve. With the cosine double-angle identity, use $\cos{2t}=1-2\sin{t}^2$ instead of the other versions so that you'll have something to factor out (the $\sin{x}$) once you get it equal to zero.

Ok for the first one, on the second part I got up to $\sin{x} = \frac {-1}{3}$, but there is now special angle that has a sine value of $\frac {-1}{3}$...what would I do next?

Then for the other one of the first equation, I got $\sin{x} = 1$, which would be $x = \frac{\pi}{2}$...would that be right? Do i have to add the $+2{\pi}k$?

And for the second problem, I got $2\sin^2{x} + \sin{x} - \frac {1}{2}$...and then I'm stuck as I can't seem to figure it out. Help?
• May 3rd 2009, 05:00 PM
chrozer
Actually I figured out whats wrong. After taking out $\cos{x}$, it should be $1+ 2\sin {x} - 4\sin^2{x}$, but this is not factorable and this is where I am stuck. Help?
• May 3rd 2009, 05:16 PM
TheEmptySet
Quote:

Originally Posted by chrozer
Actually I figured out whats wrong. After taking out $\cos{x}$, it should be $1+ 2\sin {x} - 4\sin^2{x}$, but this is not factorable and this is where I am stuck. Help?

As I did in my above post

if you factor out the 4 and complete the square you get

$-4\cos(x) [ \left(\sin(x)-\frac{1}{4} \right)^2-\frac{5}{16}]=0$

Now setting the inside =0 we get

$\left(\sin(x)-\frac{1}{4} \right)^2-\frac{5}{16}=0$

Solving for sine you get

$\sin(x)=\frac{1 \pm \sqrt{5}}{4}$

I hope this helps
• May 4th 2009, 03:18 AM
chrozer
Would the answer then just be the arcsine of $\frac{1 \pm \sqrt{5}}{4}
$
• May 4th 2009, 05:54 AM
TheEmptySet
Quote:

Originally Posted by chrozer
Would the answer then just be the arcsine of $\frac{1 \pm \sqrt{5}}{4}
$

yes