Using the identity sin^2(x) + cos^2(x) = 1sin2(x)+cos2(x)=1 we have
2cos^2(x)+sin(x)-2 = 2(1-sin^2(x))+sin(x)-22cos2(x)+sin(x)−2=2(1−sin2(x))+sin(x)−2
=2-2sin^2(x)+sin(x)-2=2−2sin2(x)+sin(x)−2
=-2sin^2(x)+sin(x)=−2sin2(x)+sin(x)
=sin(x)(-2sin(x)+1) = 0=sin(x)(−2sin(x)+1)=0
Thus the equation is true when either
sin(x) = 0sin(x)=0
or
-2sin(x)+1 = 0 <=> sin(x) = 1/2−2sin(x)+1=0⇔sin(x)=12
On the interval [0, 2pi][0,2π] we have sin(x) = 0sin(x)=0 if and only if x in {0, pi, 2pi}x∈{0,π,2π}
On the interval [0, 2pi][0,2π] we have sin(x) = 1/2sin(x)=12 if and only if x in {pi/6, (5pi)/6}x∈{π6,5π6}
Thus the set of solutions to 2cos^2(x)+sin(x)-2 = 02cos2(x)+sin(x)−2=0 is {0, pi/6, (5pi)/6, pi, 2pi}{0,π6,5π6,π,2π}
