SOLUTION: solve the multi-angle equation below sin(2x)=sq roots 2/2 I appreciate.

Question 346912: solve the multi-angle equation below
sin(2x)=sq roots 2/2 I appreciate.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
(Note: The solution below assumes that since your equation uses x nistaed of thetha for and angle, that the angles are measured in radians. If this is not correct, replace all instances of below with 180 and simplify.)

A sin value of tells you that the reference angle is
The fact that the sin value is positive tells you that the angles terminate in the quadrants where sin is positive: Quadrants I and II.
So any angle that terminates in Quadrant I or Quadrant II and has a reference angle of will have a sin of

The tricky part is expressing these angles. There are literally an infinite number of angles which fit the description above. What we do is express one angle for each quadrant and add "" (where n is any integer). This last part is how we "list" all the angles which are coterminal with the one we named.

For our Quadrant I angle we will use: itself. For our Quadrant II and we will use . Then we set the argument of sin equal to these expressions:
(where n is any integer)
and
(where n is any integer)

Last of all we solve these for x. In these equations we just have to divide both sides by 2:
(where n is any integer)
and
(where n is any integer)
These two equations together are the solution for . Replace the "n" of either equation with any integer and simplify and you will have an angle that fits the equation.
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