SOLUTION: The question I am given is "Find all values of x in the interval [0,2pi], that satisfy sin2x=cosx" I am having some difficulty with understanding why x=2pi/6 and x=4pi/6 are not i

Question 1058525: The question I am given is "Find all values of x in the interval [0,2pi], that satisfy sin2x=cosx" I am having some difficulty with understanding why x=2pi/6 and x=4pi/6 are not included in the solution since these angles are reference angles of sin(pi/6) = 1/2, and therefore, should equal to 1/2. But I just computed these angles on my calculator, and they do not equal 1/2...why is that? Thank you in advance!
Answer by ikleyn(52794) (Show Source): You can put this solution on YOUR website!
.
The question I am given is "Find all values of x in the interval [0,2pi], that satisfy sin2x=cosx" I am having some difficulty
with understanding why x=2pi/6 and x=4pi/6 are not included in the solution since these angles are reference angles of sin(pi/6) = 1/2,
and therefore, should equal to 1/2. But I just computed these angles on my calculator, and they do not equal 1/2...why is that?
Thank you in advance!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let us consider the angle x = , the first of the two angles you are asking for.

Notice that x =  = .

Now,  =  = .

From the other side,  =  = .

So, you see that sin(2x) =/= cos(x) in this case.
Therefore,  is not included to the list of solutions.
Because it IS NOT a solution.

For many other similar solved problems see the lessons
    - Solving simple problems on trigonometric equations
    - Solving typical problems on trigonometric equations
    - Solving more complicated problems on trigonometric equations
    - Solving advanced problems on trigonometric equations
in this site.

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