Another easy?trig question

  • Thread starter zeion
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  • #1
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Homework Statement



Solve the following equations of inequalities in the interval [0, 2pi)

cosx + 1 = sinx


Homework Equations





The Attempt at a Solution



cos2x + 2cosx + 1 = sin2x
cos2x + 2cosx + 1 = 1 - cos2x
2cos2x + 2cosx = 0
(2cosx)(cosx + 1) = 0
2cosx = 0 or cosx + 1 = 0
2cosx = 0
cosx = 0
x = pi/2, 3pi/2

cosx + 1 = 0
cosx = -1
x = pi

Did I miss anything?
 

Answers and Replies

  • #2
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You forgot to check your answers. When squaring an equation, you create the possibility of extraneous solutions appearing. In the case of your problem, 3pi/2 does not work in the original equation and should be left out of the solution set.
 
  • #3
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yeh you can't be too careful squaring both sides as you risk getting extra solutions, perhaps a 'cleaner' way to do it is simply to rearrange giving;

sinx - cosx = 1

then proceed to right it in the form Rsin(A-B)= 1

but your method works, like mrko said just remember to check
 

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