Question 1206160
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Solve for (0 <= x <= 360)
sin2x-cosx=-1+2sinx
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<pre>
          Step by step


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

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

cos(x)*(2sin(x) - 1) = -1 + 2sin(x)

cos(x)*(2sin(x) - 1) - (-1 + 2sin(x)) = 0

(cos(x)-1)*(2sin(x)-1) = 0


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


Case 2:  2sin(x) -1 = 0  --->  2sin(x) = 1  --->  sin(x) = 1/2  --->  x = 30 degrees or x = 150 degrees.


<U>ANSWER</U>.  There are 3 (three) solutions: x= 0 degrees;  x = 30 degrees and x= 150 degrees.
</pre>

Solved (correctly).


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For this problem, and for many other problems, &nbsp;if the restriction for the argument

is for one period, &nbsp;it is usually &nbsp;(ALWAYS) &nbsp;should be written in the form &nbsp;&nbsp;0 <= x < 360 
and &nbsp;NOT &nbsp;in the form  &nbsp;&nbsp;0 <= x <= 360, &nbsp;because &nbsp;0 &nbsp;and &nbsp;360 &nbsp;represents 
the same geometric angle &nbsp;(although different numbers).



This small detail - how do you write this restriction - in correct form &nbsp;&nbsp;0 <= x < 360

or in incorrect form &nbsp;0 <= x <= 360, &nbsp;makes it clear from the first glance

whether you know the subject and are familiar with standard mathematical notations - or not.