Question 282663
<pre><font size = 4 color = "indigo"><b>
You left off the angle for the cosine, so I can't tell whether
you meant the equation to be this: 

{{{4sin(2theta)-3cos(theta)=0}}} 

or this:

{{{4sin(2theta)-3cos(2theta)=0}}}

So I'll do it both ways:

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

If it is the first way,

{{{4sin(2theta)-3cos(theta)=0}}}

Use the identity {{{sin(2theta)=2sin(theta)cos(theta)}}}

{{{4(2sin(theta)cos(theta))-3cos(theta)=0}}}

{{{8sin(theta)cos(theta)-3cos(theta)=0}}}

Factor out {{{cos(theta)}}}

{{{cos(theta)(8sin(theta)-3)=0}}}

Use the zero-factor principle.

Set the first factor = 0:

{{{cos(theta)=0}}}

The only angles between 0° and 360° which have their
cosine equaling to zero are 90° and 270°.  So those
are two of the solutions.

Set the second factor = 0:

{{{8sin(theta)-3=0}}}

{{{8sin(theta)=3}}}

{{{sin(theta)=3/8}}}

With a calculator we find the inverse sine of {{{3/8}}}
is the first quadrant angle 22.02431284°.  However the
angle in the second quadrant whioch has 22.02431284° as
its reference angle is 157.9756872°.

So the four solutions on the interval {{{"0°"<=theta<"360°"}}},
with the decimals rounded to the nearest tenth of a degree are:

{{{theta="0°"}}}, {{{theta="270°"}}}, {{{theta="22.0°"}}}, {{{theta="158.0°"}}} 

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

If it was supposed to have been the second way:

{{{4sin(2theta)-3cos(2theta)=0}}}

{{{4sin(2theta)=3cos(2theta)}}}

Divide both sides by {{{cos(2theta)}}}

{{{4sin(2theta)/cos(2theta)=3}}}

Divide both sides by 4:

{{{sin(2theta)/cos(2theta)=3/4}}}

Use the identity {{{sin(alpha)/cos(alpha)=tan(alpha)}}}

{{{tan(2theta)=3/4}}}

With a calculator we find the inverse tangent of {{{3/4}}}
is the first quadrant angle 36.86989765°.  However the
angle in the third quadrant which has 36.86989765° as
its reference angle is 216.8698976°.

Now since the angle is {{{2theta}}} and not just {{{theta}}},
we must find all solutions for {{{2theta}}} in twice the interval, 
that is the interval {{{"0°"<=2theta<"720°"}}}, so that {{{theta}}}
will be in the interval {{{"0°"<=theta<"360°"}}}.   

So we must add 360° to each of those two values, so that when
we find {{{theta}}} by dividing by 2 we will have all the
solutions in the interval {{{"0°"<=theta<"360°"}}}.

So we have

{{{2theta="36.86989765°"}}}, {{{2theta="216.86989765°"}}}, {{{2theta="396.86989765°"}}}, {{{2theta="576.86989765°"}}},

Now solving for {{{theta}}}, we have:

{{{theta="18.43494882°"}}}, {{{theta="108.4349488°"}}}, {{{theta="198.4349488°"}}}, {{{theta="288.4349488°"}}},

Or rounded to the nearest tenth of a degree:

{{{theta="18.4°"}}}, {{{theta="108.4°"}}}, {{{theta="198.4°"}}}, {{{theta="288.4°"}}}.

Edwin</pre>