Question 1177378
<pre>
It cannot be true for {{{x=pi/6=30^o}}} or {{{x=5pi/6=150^o}}} or any angle
which is that plus or minus a multiple of 2&pi; or 360°. That's because
there will be a zero denominator then.  You might point that out to your
teacher that the left side is undefined in those cases.

Here's the way to find a and b for all other values of x.

 {{{ ( -3 +  4cos^2(x)) / ( 1 - 2sin^""(x) ) = a + b*sin^""(x) }}}

It has to be true when x=0.

So let's plug in 0 for x:

{{{ ( -3 +  4cos^2(0)) / ( 1 - 2sin^""(0) ) = a + b*sin^""(0) }}}

{{{ ( -3 +  4(1)^2 ) / ( 1 - 2(0) ) = a + b*(0) }}}

{{{ ( -3 +  4 ) / ( 1 ) = a  }}}

{{{1/1=a}}}

{{{1=a}}}

Substitute 1 for a in

 {{{ ( -3 +  4cos^2(x)) / ( 1 - 2sin^""(x) ) = a + b*sin^""(x) }}}

 {{{ ( -3 +  4cos^2(x)) / ( 1 - 2sin^""(x) ) = 1 + b*sin^""(x) }}}

It also must be true when x = 90° or &pi;/2

 {{{ ( -3 +  4cos^2(90^o)) / ( 1 - 2sin^""(90^o) ) = 1 + b*sin^""(90^0) }}}

 {{{ ( -3 +  4(0)) / ( 1 - 2(1) ) = 1 + b*(1) }}}

 {{{ (-3) / ( 1 - 2) = 1 + b }}}

{{{(-3)/(-1)=1+b}}}

{{{3=1+b}}}

{{{2=b}}}

So a=1 and b=2

Edwin</pre>