Hardly seen question (don't know if it exists)

[NeedAspirin]

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?

[Special At Specialist]

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))

[tony3272]

For this case though you would provide a general solution for and

For :



For :

  *Edit* forget this line since it's wrong (I forgot that it's negative) go to next

or


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

[dc302]

For this case though you would provide a general solution for and

For :



For :




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

No real solution at least.

[NeedAspirin]

Thanks, for answering!