# help with solving for x

• Aug 20th 2013, 11:49 PM
mikeythemaniac
help with solving for x
the problem im having trouble with is: solve 1 + cos(2x) = 2sin^2(2x) for x, where 0 ≤ x < 2π

i set everything equal to 0 and worked it down to (cos(2x) + 1)(2cos(2x) - 1) = 0 and solved each of those individually getting x = π/2 and x = π/6 but I'm not even sure if everything is correct to that point or how to finish the problem off from there if I was going on the right direction. any help would be much appreciated. thanks
• Aug 20th 2013, 11:57 PM
chiro
Re: help with solving for x
Hey mikeythemanic.

After you verify by plugging in, see if you have other values that lie in the [0,2pi) interval. Remember that cos(x) = cos(2*pi-x) and sin(x) = -sin(2*pi-x).
• Aug 21st 2013, 12:00 AM
Re: help with solving for x
You are missing the values of 5pi/6 and 3pi /2.

What really helps with this problem is to set 2x = z.
The only identity you will need to know is that sin^2 x + cos^2 x = 1. Use that to get rid of the sin in the equation.

When you plug 2x back in and you have something like 2x = pi, remember to do this 2x = pi + 2pi*n (n being any integer). Remember that 2x provides double the solutions as x, so 2x = pi + 2pi*n becomes x = pi/2 +pi*n, the range is [0,2pi) so will have pi/2 and 3pi/2.
• Aug 21st 2013, 12:26 AM
mikeythemaniac
Re: help with solving for x
i ended up with cosx = 2sinxcosx and divide by cosx, then again by 2 and i get sinx = 1/2

so back to the original problem, would i then just say x = arcsin(1/2) or would it be pi/6?

and 1 last question, how am i so retarded? i feel like it's my gift.
• Aug 21st 2013, 12:27 AM
mikeythemaniac
Re: help with solving for x
ok now im really confused since you edited your post, was i doing it the right way before? my head is going to explode soon i think.
• Aug 21st 2013, 01:01 AM
Re: help with solving for x
Let me walk you through the problem.

From the original equation, I set 2x = z.

1 + cosz - 2sin^2 z = 0
1 + cosz - 2(1-cos^2 z) = 0 Remember sin^2 x + cos^2 x = 1
2cos^2 z + cosz - 1 = 0
(2cosz - 1) (cosz + 1) = 0
Starting with 2cosz - 1 =0
**For problems with a coefficient next to x, set the roots up as if there were no restraints to values of x
cosz = 1/2
z = pi/3 + 2pi*n or z = 5pi/3 + 2pi*n **n being any value since this statement represent all possible values with out any range
2x = pi/3 + 2pi*n or 2x = 5pi/3 + 2pi*n
x = pi/6 + pi*n or x = 5pi/6 + pi*n
x = pi/6 or 7pi/6 or x = 5pi/6

now with cosz + 1 = 0

cosz = -1
z = pi + 2pi*n
2x = pi + 2pi*n
x = pi/2 + pi*n
x = pi/2 or 3pi/2

In all, our values are pi/6, 7pi/6, 5pi/6, pi/2, 3pi/2.

Understand why? i left so values out of my previous post
• Aug 21st 2013, 01:31 AM
mikeythemaniac
Re: help with solving for x
i think im starting to understand it better, so my final answer would be those 5 answers because the range given for x was between 0 and 2pi? getting this many answers as an answer is a bit new to me, gonna take some getting used to i guess. thanks for walking me through it, helps a lot.
• Aug 21st 2013, 01:49 AM
Re: help with solving for x
My apologies, I forgot to include 11pi/6 in the final answer.
I forgot about it in the equation x = 5pi/6 + pi*n. Both 5pi/6 and 11pi/6 are possible
• Aug 21st 2013, 02:06 AM
mikeythemaniac
Re: help with solving for x
no worries, im certainly not good enough at this to catch many mistakes. lol

thanks again. :)