Question 823002
Note: After seeing your thank you note I came back to my solution and checked. I'm sorry to say that I was wrong about the reference angle. A reference angle of {{{pi/6}}} not {{{pi/3}}} has a sin of 1/2. Everything else was correct. The problem below has been redone using the proper reference angle.<br>
{{{2sin(2theta) = -1}}}
First isolate the function and its argument by dividing both sides by 2:
{{{sin(2theta) = -1/2}}}<br>
We should recognize that 1/2 (positive or negative) is a special angle value for sin. Without our calculator we should know that a 1/2 for sin indicates a reference angle of {{{pi/6}}}. Since the 1/2 is negative and since sin is negative in the 3rd and 4th quadrants we should get the following general solution equations (for 2 theta):
{{{2theta = pi+pi/6+2pi*n}}} (for the 3rd quadrant)
{{{2theta = -pi/6+2pi*n}}} (for the 4th quadrant)
The first equation will simplify:
{{{2theta = 7pi/6+2pi*n}}}
{{{2theta = -pi/6+2pi*n}}}
Dividing both sides of both equations by 2:
{{{theta = 7pi/12+pi*n}}}
{{{theta = -pi/12+pi*n}}}<br>
Now we try various integer values for n as we search for specific solutions which are in the specified interval.
From {{{theta = 7pi/12+pi*n}}}:
If n = 0 then {{{theta = 7pi/12}}}
If n = 1 then {{{theta = 19pi/12}}}
If n = 2 (or larger) then {{{theta}}} is too large for the interval
If n = -1 (or smaller) then {{{theta}}} is too small for the interval
From {{{theta = -pi/12+pi*n}}}:
If n = 0 (or smaller) then {{{theta}}} is too small for the interval
If n = 1 then {{{theta = 11pi/12}}}
If n = 2 then {{{theta = 23pi/12}}}
If n = 3 (or larger) then {{{theta}}} is too large for the interval<br>
So the only solutions in the specified interval are: {{{7pi/12}}}, {{{19pi/12}}}, {{{11pi/12}}} and {{{23pi/12}}}