Question 1042988
<pre><font size = 4><b>The good lady did not get all the solutions
between 0 and {{{2pi}}}.  There are 6 solutions for
the first one and 4 solutions for the second one.
</pre>
How can i solve these trigonometric functions? 
Solve for x&#8712;[0,2&#960;]:
{{{sin(3x)=-1/sqrt(2)}}}
<pre>
The sine is negative in QIII and QIV

Rationalize the denominator of {{{-1/sqrt(2)}}}
and get {{{-sqrt(2)/2}}} and from your knowledge
of special angles or from the unit circle,
the solutions for 3x are 225° and 315° or
since you are using radians, {{{5pi/4}}} and {{{7pi/4}}}.

However, the left side is sin(3x), not sin(x).

That means we must get 3 times as many answers for x
as we would get for 3x.

To do that we add {{{2pi}}} 3-1 = 2 times to each of those
answers for 3x in order to get all non-negative answers for 
x less than {{{2pi}}}:

So 

1.  we have to add {{{2pi}}} and {{{4pi}}} to the first answer{{{5pi/4}}}:  

{{{matrix(1,9,3x,"",""="","",5pi/4,",",5pi/4+2pi,",",5pi/4+4pi)}}}

which we get LCDs and simplify as

{{{matrix(1,9,3x,"",""="","",5pi/4,",",5pi/4+8pi/4,",",5pi/4+16pi/4)}}}

and simplify further as

{{{matrix(1,9,3x,"",""="","",5pi/4,",",13pi/4,",",21pi/4)}}}

Then divide through by 3 (multiply through by {{{1/3}}} and get:

{{{matrix(1,9,3x(1/3),"",""="","",(5pi/4)(1/3),",",(13pi/4)(1/3),",",(21pi/4)(1/3))}}}

or

{{{matrix(1,9,x,"",""="","",5pi/12,",",13pi/12,",",7pi/4)}}}

Also,

2.  we have to add {{{2pi}}} and {{{4pi}}} to the second answer{{{7pi/4}}}:  

{{{matrix(1,9,3x,"",""="","",7pi/4,",",7pi/4+2pi,",",7pi/4+4pi)}}}

We get the LCD:

{{{matrix(1,9,3x,"",""="","",7pi/4,",",7pi/4+8pi/4,",",7pi/4+16pi/4)}}}

and simplify further as

{{{matrix(1,9,3x,"",""="","",7pi/4,",",15pi/4,",",23pi/4)}}}

Then divide through by 3 (multiply through by {{{1/3}}}) and get:

{{{matrix(1,9,3x(1/3),"",""="","",(7pi/4)(1/3),",",(15pi/4)(1/3),",",(23pi/4)(1/3))}}}

or

{{{matrix(1,9,x,"",""="","",7pi/12,",",5pi/4,",",23pi/12)}}}

All the 6 solutions in order of magnitude are

{{{matrix(1,15,x,"",""="","",5pi/12,",",7pi/12,",",13pi/12,",",5pi/4,",",7pi/4,",",23pi/12)}}}

------------------------

Solve for x&#8712;[0,2&#960;]:

{{{sin(2x)=-1/2}}}

As before, the sine is negative in QIII and QIV.

From your knowledge of special angles or from the unit 
circle, the solutions for sin(2x) are 210° and 330° or
since you are using radians, {{{7pi/6}}} and {{{11pi/6}}}.

However, the left side is sin(2x), not sin(x).

That means we must get 2 times as many answers for x
as we would get for 2x.

To do that we add {{{2pi}}} 2-1 = 1 time to each of those
answers for 2x in order to get all non-negative answers for 
x less than {{{2pi}}}:

So 

1.  we have to add {{{2pi}}} to the first answer{{{7pi/6}}}:  

{{{matrix(1,7,2x,"",""="","",7pi/6,",",7pi/6+2pi)}}}

We get the LCD:

{{{matrix(1,7,2x,"",""="","",7pi/6,",",7pi/6+12pi/6)}}}

and simplify further as

{{{matrix(1,7,2x,"",""="","",7pi/6,",",19pi/6)}}}

Then divide through by 2 (multiply through by {{{1/2}}}) and get:

{{{matrix(1,7,2x(1/2),"",""="","",(7pi/6)(1/2),",",(19pi/6)(1/2))}}}

or

{{{matrix(1,7,x,"",""="","",7pi/12,",",19pi/12)}}}

Also,

2.  we have to add {{{2pi}}} to the second answer {{{11pi/6}}}:  

{{{matrix(1,7,2x,"",""="","",11pi/6,",",11pi/6+2pi)}}}

which we get LCDs and simplify as

{{{matrix(1,7,2x,"",""="","",11pi/6,",",11pi/6+12pi/6)}}}

and simplify further as

{{{matrix(1,7,2x,"",""="","",11pi/6,",",23pi/6)}}}

Then divide through by 2 (multiply through by {{{1/2}}} and get:

{{{matrix(1,7,2x(1/2),"",""="","",(11pi/6)(1/2),",",(23pi/6)(1/2))}}}

or

{{{matrix(1,7,x,"",""="","",11pi/12,",",23pi/12)}}}


All the 4 solutions in order of magnitude are

{{{matrix(1,11,x,"",""="","",7pi/12,",",11pi/12,",",19pi/12,",",23pi/12)}}} 

Edwin</pre></b>