SOLUTION: Find all solutions of the equation in the interval [0, 2 pi) -sin2x+2cosx=0

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Question 979670: Find all solutions of the equation in the interval
[0, 2 pi)
-sin2x+2cosx=0
Answer by ikleyn(52765) About Me  (Show Source):
You can put this solution on YOUR website!

The equation is

2cosx = sin2x.                      (1)

Apply the formula for sines of the double argument:  sin2x = 2sinx%2Acosx  (see,  for example,  the lesson  Trigonometric functions of multiply argument  in this site).
Then the equation takes the form

2cosx = 2%2Acosx%2Asinx.             (2)

One solution of this equation is  cosx = 0,  which gives  x = pi%2F2,  3pi%2F2  (regarding the interval [0, 2pi) ).

Further,  let us assume that  cosx  is not equal to zero.  Then we can reduce the equation  (2) dividing its both sides by  2%2Acosx.  You will get

1 = sinx,

which gives the solutions  x = pi%2F2  we just obtained above.

Answer.  The solutions are  x = pi%2F2,  3pi%2F2.