SOLUTION: I need help figuring out all the solutions in the interval [0,2pi) sin^2(x)+2cos(x)=2

Question 1206254: I need help figuring out all the solutions in the interval [0,2pi)
sin^2(x)+2cos(x)=2
Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
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I need help figuring out all the solutions in the interval [0,2pi)
sin^2(x)+2cos(x)=2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The idea is to replace sin^2(x) by expression which contains cos(x), only.
        Then the entire equation will be relative cos(x), and we will be able solve it.

So, by implementing this idea, we write  sin^2(x) = 1 - cos^2(x).

Then the given equation takes the form

    (1-cos^2(x)) + 2cos(x) = 2.


Simplify it

    - cos^2(x) + 2cos(x) = 1

    cos^2(x) - 2cos(x) + 1 = 0

     = 0   --->

    cos(x) = 1  --->  x = 0.


ANSWER.  In the given interval, the original equation has a UNIQUE root  x = 0 radians.

Solved.

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