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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: NeedAspirin on November 06, 2011, 09:13:10 pm

Title: Hardly seen question (don't know if it exists)
Post by: NeedAspirin on November 06, 2011, 09:13:10 pm: Is it possible to have a question like this or similar to this:

cos^2(x)=a (a is just some random number)

If so, how do you solve it?
Title: Re: Hardly seen question (don't know if it exists)
Post by: Special At Specialist on November 06, 2011, 09:17:46 pm: cos^2(x) = a
Take the +or- square root of both sides.
cos(x) = +or- sqrt(a)
Now take the arccos of both sides:
x = arccos(sqrt(a)) or x = arccos(-sqrt(a))
In other words, the solutions between the interval [0, 2pi] are:
x = arccos(sqrt(a)), x = pi - arccos(sqrt(a)), x = pi + arccos(sqrt(a)) and x = 2pi - arccos(sqrt(a))
Title: Re: Hardly seen question (don't know if it exists)
Post by: tony3272 on November 06, 2011, 09:26:54 pm: For this case though you would provide a general solution for $cos(x)=\sqrt{a}$ and $cos(x)=-\sqrt{a}$

For $cos(x)=\sqrt{a}$ :

$x=2n\pi\pm cos^{-1}(\sqrt{a})$

For $cos(x)=-\sqrt{a}$ :

$x=2n\pi\pm cos^{-1}(-\sqrt{a})$ *Edit* forget this line since it's wrong (I forgot that it's negative) go to next

$x=2n\pi-\frac{\pi}{2}- sin^{-1}(-\sqrt{a})$ or $x=2n\pi+\frac{\pi}{2}+sin^{-1}(-\sqrt{a})$

This is assuming a is between 0 and 1, otherwise there is no solution.
Title: Re: Hardly seen question (don't know if it exists)
Post by: dc302 on November 06, 2011, 09:33:55 pm: Quote from: tony3272 on November 06, 2011, 09:26:54 pm
For this case though you would provide a general solution for $cos(x)=\sqrt{a}$ and $cos(x)=-\sqrt{a}$

For $cos(x)=\sqrt{a}$ :

$x=2n\pi\pm cos^{-1}(\sqrt{a})$

For $cos(x)=-\sqrt{a}$ :

$x=2n\pi\pm cos^{-1}(-\sqrt{a})$

This is assuming a is between 0 and 1, otherwise there is no solution.

No real solution at least. :P
Title: Re: Hardly seen question (don't know if it exists)
Post by: NeedAspirin on November 06, 2011, 09:43:31 pm: Thanks, for answering!