Question 332270
how do i find all the intervals of x which satisfy 2cos2x = 5 -5sinx, in radians?
<pre><font size = 4 color = "indigo"><b>
{{{2*Cos(2x) = 5 -5*Sin(x)}}}

Use the identity {{{Cos(2theta)=1-2Sin^2theta}}} 

{{{2(1-2*Sin^2(2x)) = 5 -5*Sin(x)}}}

{{{2-4*Sin^2(2x)=5-5*Sin(x)}}}

{{{-4*Sin^2(2x)+5*Sin(x)-3=0}}}

Multiply through by -1:

{{{4*Sin^2(2x)-5*Sin(x)+3=0}}}

That is a quadratic in {{{Sin(x)}}}.  However

the discriminant is {{{b^2-4ac=(-5)^2-4(4)(3)=25-48=-23}}}

Since the discriminant is negative, the solutions to the quadratic are
imaginary.

Therefore there can be no real number solutions.

Edwin</pre>