Question 167218
For the life of me, I cannot solve 4cos^2x - sinx -5 = 0. Any help would be appreciated

{{{4COS^2x - SINx -5 = 0}}}
<pre><font size = 4 color = "indigo"><b>
Use the Pythagorean identity {{{COS^2alpha=1-SIN^2alpha}}}
to replace {{{COS^2x}}} by {{{1-SIN^2x}}}

{{{4(1-SIN^2x) - SINx -5 = 0}}}

{{{4-4SIN^2x - SINx -5 = 0}}}

{{{-4SIN^2x - SINx -1 = 0}}}

Multiply thru by {{{-1}}}

{{{4SIN^2x + SINx +1 = 0}}}

This is a quadratic in {{{SINx}}}

But it can have no solution since the discriminant
would be negative  {{{b^2-4ac=1^2-4*4*1=1-16=-15}}}

Edwin</pre>