Question

Find the general solution of 2cos^2(x) =1

Answer 1

\[2\cos^2 (x) = 1\]\[\cos^2(x) = \frac{1}{2}\]Look at your unit circle — where is that true? Do you see a pattern? Don't forget to figure out where it is true for negative values of \(x\), too.

Answer 2

But how do you get it in this form?
\[\theta=360n \pm \alpha\]

Answer 3

where n is an integer and \[\cos \theta = \cos\]

Answer 4

first, why don't you tell me the 4 spots on the unit circle that satisfy the equation...

Answer 5

okay well x= 45, 135, 225, 315
is that correct?

Answer 6

I would normally do this in terms of radians, where the answers would be \[\pi/4, -\pi/4, 3\pi/4, -3\pi/4\]

Once we find a formula, we can always convert the amounts to degrees if that's what you really need

Answer 7

looks to me like that is \[\pm(\frac{\pi}{4} + 2\pi n), n\in \text{Integers} \] and \[\pm(\frac{3\pi}{4}+2\pi n), n\in\text{Integers}\]

Answer 8

I haven't learnt radians though, is there a simpler method?

Answer 9

or just \[\frac{\pi}{4} \pm \frac{\pi}{2}n\]

\[\frac{\pi}{4} * \frac{360^\circ}{2\pi} = 45^\circ\]\[\frac{\pi}{2} * \frac{360^\circ}{2\pi} = 90^\circ\]

So your angles are just \(45^\circ \pm 90^\circ n\) where \(n\) is an integer

and here's a table of the values from n = -10 to 10

\[\begin{array}{cc}
x & \cos^2(x)\\\hline\\
-855 {}^{\circ} & \frac{1}{2} \\
-765 {}^{\circ} & \frac{1}{2} \\
-675 {}^{\circ} & \frac{1}{2} \\
-585 {}^{\circ} & \frac{1}{2} \\
-495 {}^{\circ} & \frac{1}{2} \\
-405 {}^{\circ} & \frac{1}{2} \\
-315 {}^{\circ} & \frac{1}{2} \\
-225 {}^{\circ} & \frac{1}{2} \\
-135 {}^{\circ} & \frac{1}{2} \\
-45 {}^{\circ} & \frac{1}{2} \\
45 {}^{\circ} & \frac{1}{2} \\
135 {}^{\circ} & \frac{1}{2} \\
225 {}^{\circ} & \frac{1}{2} \\
315 {}^{\circ} & \frac{1}{2} \\
405 {}^{\circ} & \frac{1}{2} \\
495 {}^{\circ} & \frac{1}{2} \\
585 {}^{\circ} & \frac{1}{2} \\
675 {}^{\circ} & \frac{1}{2} \\
765 {}^{\circ} & \frac{1}{2} \\
855 {}^{\circ} & \frac{1}{2} \\
945 {}^{\circ} & \frac{1}{2} \\
\end{array}\]

Answer 10

it's late, I'm not explaining things very well, time to call it a night!

Answer 11

Hi,

Just to continue where @whpalmer4 has left off (although I think that @whpalmer4 has done a pretty good job here though).

From @whpalmer4 's previous post:

\[\cos ^{2}x = \frac{ 1 }{ 2 } \]
\[\cos x = \pm \sqrt{\frac{ 1 }{ 2 }}\]

x = arccos ( \sqrt{\frac{ 1 }{ 2 }}\ ) = 45 deg

General solution for x: \[\theta = 360n \pm 45\], where \[n \in \mathbb{Z}^* \].

Answer 12

Yes, I believe that the general solution for x is \[\theta =360\pm45\]
but apparently the textbook answer is \[\theta =180n \pm 45\]
So i am quite confused.

Answer 13

@fishy13 : Oh yes, I forgot that cosx = +/- sqrt (1/2), which means that x can lie in all 4 quads. So the correct answer should be \[\theta = 180n \pm 45\].

Answer 14

And actually \[\theta = 90n\pm 45^\circ\]is equivalent, although it supplies each point twice...

Answer 15

In any case, @shiraz14, thanks for picking up where I left off! Feel free to do so anytime you see the need...

Answer 16

@whpalmer4 : You're welcome.
@fishy13 : Yes, this is a special case where you can also have 90n +/- 45 degrees (as @whpalmer4 has stated) simply because 45 degrees is the bisector of 90 degrees. But be careful that this does not always apply, since if theta = 30 degrees, etc, only the conventional argument will work.