Recall the Pythagorean identity,
[tex]\sin^2x+\cos^2x=1[/tex]
So the equation is the same as
[tex](1-\cos^2x)+\cos x=2\implies\cos^2x-\cos x+1=0[/tex]
Complete the square:
[tex]\cos^2x-\cos x+\dfrac14+\dfrac34=0[/tex]
[tex]\left(\cos x-\dfrac12\right)^2+\dfrac34=0[/tex]
[tex]\implies\left(\cos x-\dfrac12\right)^2=-\dfrac34[/tex]
But whenever you square a real number, you get a positive number, so there are no real solutions to this equation.
